Evaluation of Integrals by Differentiation with Respect to a Parameter 1 Introduction and Three Examples

Home , Constant of integration, Leibniz integral rule

December 2018

Evaluation of integrals by differentiation with respect to a parameter

Khristo N. Boyadzhiev

Department of Mathematics and Statistics, Ohio Northern University, Ada, OH 45810, USA [email protected]

Abstract. We review a special technique for evaluating challenging integrals by providing a number of examples. Many of our examples prove integrals from the popular table of Gradshteyn and Ryzhik.

Keywords. Improper integrals; integrals depending on a parameter; integral evaluation; Laplace transform; Laplace integrals.

Mathematics subject classification. 26A42; 30E20; 33E20.

1 Introduction and three examples

There are various methods for evaluating integrals: substitution, integration by parts, partial fractions, using the residue theorem, or Cauchy’s integral formula, etc. A beautiful special technique is the differentiation with respect to a parameter inside the integral. We review this technique here by providing numerous examples, many of which prove entries from the popular handbook of Gradshteyn and Ryzhik [6]. In our examples we focus on the formal manipulation. Several theorems justifying the legitimacy of the work are listed at the end of the paper. Applying the theorems in every particular case is left to the reader.

We hope integral lovers will appreciate this review and many will use it as a helpful reference. The examples and techniques are accessible to advanced calculus students and can be applied in various projects. Many more integrals from [6] can be proved by using the same approach.

We also want to mention that for many of the presented examples, solving the integral by differentiation with respect to a parameter is possibly the best solution.

Our main reference is the excellent book of Fikhtengolts [5] which is the source of several examples. Some integrals solved by this technique can be found in [1] and [2]. The method is presented in various publications, for instance, [4], [7], [9], [10], [12], and [13].

Below we evaluate three very different integrals in order to demonstrate the wide scope of the method. In section 2 we present a collection of eighteen more or less typical cases. In section 3 we show how differential equations can be involved very effectively. In section 4 we demonstrate a more sophisticated technique, where the parameter appears also in the integral limits.

Example 1.1

We start with a very simple example. It is easy to show that the popular integral

∞ sin x (1.1) ∫ dx 0 x is convergent. In order to evaluate this integral we introduce the function

∞ sin x (1.2) F()λλ= ∫ e−λx dx,> 0 0 x and differentiate this function to get

∞ −1 F′(λ )=−= e−λx sin xdx ∫ 2 0 1+ λ

(Laplace transform of the sine function). Integrating back we find

FC(λλ )=−+ arctan( ) .

Setting λ →∞ yields the equation

ππ 0=−+C , ie .. C = , 22

and therefore,

∞ sin x π (1.4) F()λλ= ∫ e−λx dx = − arctan 0 x 2

(cf. entry 3.9411 in [6]). Taking limits for λ → 0 we find

∞ sin x π (1.5) ∫ dx = . 0 x 2

Example 1.2

The 66 Annual William Lowell Putnam Mathematical Competition (2005) included the integral (A5)

1 ln (1+ x ) (1.6) dx ∫ 2 0 1+ x with a solution published in [11]. This is entry 4.291(8) in [6]. We shall give a different solution by introducing a parameter. Consider the function

1 ln (1+ λx ) (1.7) F()λ = dx ∫ 2 0 1+ x

defined for λ ≥ 0 . Differentiating this function we get

1 x F′()λ = dx . ∫ 2 0 (1++λxx )(1 )

This integral is easy to evaluate by splitting the integrand in partial fractions. The result is

ln(1+ λ ) 1 1 πλ F′()λ =−++ln 2 . 1+λ2 2 1 ++ λλ 22 41

Integrating we find

λ ln(1+ x ) ln 2 π F()λ=−+dx arctanλλ ++ ln (12 ) (1.8) ∫ 2 , 0 128+ x

and setting λ =1 we arrive at the equation

π 2F (1)= ln 2 . 4

That is,

1 ln (1+ x ) π (1.9) dx = ln 2 . ∫ 2 0 18+ x

As we shall see later, many integrals containing logarithms and inverse trigonometric functions can be evaluated by this method.

It is good to notice that integrating (1.9) by parts we find

11 ln(1+ xx ) 1 arctan dx=+−ln(1 x )arctan x dx ∫∫2 0 0011++xx and therefore, we have also

1 arctan x π ∫ dx = ln 2 . 0 18+ x

Example 1.3

This is Problem 1997 from the Mathematics Magazine 89(3), 2016, p. 223. Evaluate

2 ∞ 1− e−x (1.10) ∫dx . 0 x

Solution. We show that for every λ > 0

2 ∞ 1− e−λx (1.11) F()λλ≡=∫dx ln 4 . 0 x

Indeed, differentiating this function with respect to λ (which is legitimate, as the integral is uniformly convergent on every interval 0 <<

∞∞−−λx −λλxx2 1−−e−λx ee F′(λ )= 2∫∫e dx = 2 dx = 2ln 2 00xx by using Frullani’s formula for the last equality (see below).

We conclude that F()λ is a linear function and since F(0)= 0 we can write F(λλ )= ln 4 . With λ =1 we find F(1)= ln 4 .

Frullani’s formula says that for appropriate functions fx() we have

∞ f() ax− f () bx b (1.12) ∫ dx=[ f (0) −∞ f ( )] ln . 0 xa

2. General examples

Example 2.1

We start this section with a simple and popular example Consider the integral

1 xα −1 (2.1) J ()α = ∫ dx 0 ln x

for α ≥ 0. Note that the integrand is a continuous function on [0,1] when it is defined as zero at d x = 0 and as α (its limit value) at x =1. Since xxxαα= ln , differentiation with respect to dα α yields

1 1 J′()α =∫ xα dx = 0 1+α

and JC(αα )= ln(1 ++ ) . Since J (0)= 0 we find C = 0 and finally,

1 xα −1 (2.2) ∫ dx =ln (1 +α ) . 0 ln x

The similar integral

1 xxαβ− (2.3) ∫ dx 0 ln x

where αβ,0≥ can be reduced to (2.2) by writing xxxαα−ββ = −−1 ( x − 1) . Thus we have

1 xxαβ−+1 α (2.4) ∫ dx = ln . 0 lnx 1+ β

Another way to approach (2.1) is to use the substitution xe= −t which transforms it to the Frullani integral

∞ ee−+(α 1)t − −t (2.5) ∫ dt 0 t

see (1.12).

Example 2.2

We evaluate here the improper integral

1 arctan λx (2.6) J ()λ = dx . ∫ 2 0 xx1−

Differentiation yields

1 dx J′()λ = ∫ 22 2 0 (1+−λ xx ) 1

and with the substitution x = cosθ this transforms into

ππ/2 1 /2 dθ Jd′()λθ= = ∫∫22 2 2 2 001+λ cos θ (1++ tanθλ )cos θ

π π /2 d tanθ 1 tanθπ2 = = arctan = . ∫ 22 2 22 0 1++λθ tan 1+λ 1 ++ λλ 21 0

Therefore,

π (2.7) J (λ )= ln λλ ++ 1 2 , 2 ( )

since J (0)= 0 . In particular, with λ =1,

1 arctan x π (2.8) dx =ln (1 + 2 ) . ∫ 2 0 xx1− 2

This integral is entry 4.531(12) in [6]. Note that the similar integral

1 arctan λx (2.9) J ()λ = dx ∫ 2 0 1− x

cannot be evaluated in the same manner. The derivative here becomes

1 xdx π /2 cosθ 1 1++λλ2 Jd′()λθ= = = ln ∫∫22 2 22 2 2 00(1+λxx ) 1 − 1+ λθ cos 2λ 1+ λ 1 +− λλ

which is not easy to integrate. The integral (2.9) will be evaluated later in section 4 by a more sophisticated method.

Example 2.3

Now consider

∞ ln(1+ λ 22x ) (2.10) J ()λ = dx ∫ 2 0 1+ x

with

∞ 2λx2 J′()λ = dx ∫ 22 2 0 (1++λ xx )(1 )

2λ∞  11  2λ  ππ π = −dx = −= 2∫ 22 2 2  1−+λ0  1 λxx 1 +  1 −λλ  221 + λ under the restriction λ 2 ≠ 1. This way

(2.11) J (λπ )= ln(1 + λ ) .

We needed λ 2 ≠ 1 for the evaluation of J′()λ , but this restriction can later be dropped. For equation (2.11) we only need λ >−1. In particular, for λ =1,

∞ ln(1+ x2 ) dx = π ln 2 . ∫ 2 0 1+ x

The similar integral

1 ln(1+ x22 ) 3 π ∫ dx =(ln 2)2 − 0 1+ x 4 48 is evaluated by the same method in [3]. In that article one can find also the evaluation

1 ln(1+ x2 ) π dx=ln 2 − G , ∫ 2 0 12+ x where G is Catalan’s constant (see the remark at the end of Section 4).

Example 2.4

We shall evaluate here two more integrals with arctangents. The first one is 4.535(7) from [6],

∞ arctan λx (2.12) G()λ = dx . ∫ 2 0 xx(1+ )

For all λλ>−1, ≠ 1 we compute

∞∞dx 11λ 2 G′()λ = = − dx ∫∫22 2 2 2 22 00(1+λxx )(1 + ) 1 − λλ 1 ++ x 1 x

1 π πλ π = −= , 1−+λλ2  2 2 2(1 )

and hence (dropping the restriction λ ≠ 1)

π (2.13) G(λλ )= ln (1 + ) . 2

and for λ =1

∞ arctan x π (2.14) dx = ln 2 . ∫ 2 0 xx(1+ ) 2

Comparing this to (2.11) we conclude that for all λ >−1

∞∞ln(1+ λλ22xx ) arctan (2.15) dx = 2 dx =πλln (1 + ) . ∫∫22 001 ++x xx(1 )

Example 2.5

Related to (2.12) is the following integral

∞ arctan(λµxx )arctan( ) (2.16) G(,λµ )= dx ∫ 2 0 x

for λµ,0> . Using the evaluation (2.13) we write

∞ arctan(µπλx )  G (,)λµ = dx =ln 1 + λ ∫ 22  0 xx(1+ λµ ) 2  and integrating this logarithm by parts with respect to λ we find

π GC(λµ , )= [( λµ + ) ln( λµ +− ) λλ ln ] + ( µ ) . 2

π Setting λ → 0 yields C(µ )= − µµ ln . Finally, 2

∞ arctan(λxx )arctan( µπ ) (2.17) dx =[(λµ + ) ln( λµ +− ) λλµ ln − ln µ ] . ∫ 2 0 x 2

Example 2.6

Consider the integral 3.943 from [6]

∞ 1− cos λx F()λ = ∫ e−β x dx, 0 x

where β > 0 is fixed. We have

∞ λ F′(λλ )= e−β x sin xdx = ∫ 22 0 λβ+

and integrating back

1 FC(λ )= ln( λβ22 ++ ) ( β ) . 2

−1 To compute C()β we set λ = 0 and this gives C(ββ )= ln 2 . Therefore, 2

∞ 1− cosλλx 1 2 e−β x dx =ln 1 + (2.18) ∫ 2 . 0 x 2 β

Example 2.7

A “symmetrical” analog to the previous example is the integral

∞ 1− e−λx F(λβ )= ∫ cosxdx , 0 x

defined for λ ≥ 0 and β ≠ 0 . The integral is divergent at infinity when β = 0 . We have

∞ λ F′(λβ )= e−λ x cos xdx = ∫ 22 0 λβ+ and integrating

∞ 11− e−λx λ 2 F(λβ )= cosxdx = ln 1 + (2.19) ∫ 2 0 x 2 β

so that for any λβ≥>0, 0

∞∞1−−ex−λx 1 cos λ ∫∫cos β xdx= e−β x dx . 00xx

Note that the integral 3.951(3) from [6]

∞ ee−−λxx− µ ∫ cos β xdx 0 x

can be reduced to (2.19) by writing ee−−λλxx−µµ =( e − x −+− 1) (1 e− x ) and splitting it in two integrals. Thus

∞ ee−−λxx−+µ 1 µβ22 cosβ xdx = ln . ∫ 22 0 x 2 λβ+

Example 2.8

Using the well-known Gaussian integral, also known as the Euler-Poisson integral,

∞ 2 π (2.20) ∫ e−x dx= 0 2 we can evaluate for every λ ≥ 0 the integral

2 ∞ 1− e−λx F()λ = dx . ∫ 2 0 x

Indeed, we have for λ > 0

∞∞ 2 1 π F′()λ==−=∫∫ e−λx dx exp( (xλλ )2 ) d x , 00λλ2 so that

2 ∞ 1− e−λx (2.21) F()λ = dx = λπ . ∫ 2 0 x

Example 2.9

Sometimes we can use partial derivatives as in the following integral. Consider the function

∞ e−−px cos qx− eλ x cos µ x (2.22) F(,λµ )= ∫ dx 0 x with four parameters. Here λµ> 0, will be variables and pq> 0, will be fixed. The partial derivatives are

∞ λ F(λµ , )= e−λ x cos µ xdx = , λ ∫ 22 0 λµ+

∞ µ F(λµ , )= e−λ x sin µ xdx = . µ ∫ 22 0 λµ+

It is easy to restore the function from these derivatives

1 F(λµ , )= ln( λ22 ++ µ )C ( pq , ) , 2 where C(,) pq is unknown. The integral (2.22) is zero when λ = p and µ = q , so from the last equation we find C( pq , )=−+ ln( p22 q ) / 2 . Therefore,

∞ e−−px cos qx−+ eλ x cosµ x 1 λµ22 (2.23) dx = ln . ∫ 22 0 x 2 pq+

In all following examples containing “ e−λx ” we assume λ > 0 .

Example 2.10

Now consider

∞ sin(ax )sin( bx ) J()λ = ∫ e−λx dx 0 x

where ab>>0 are constants. Clearly,

∞ J′(λ )= −∫ e−λx sin( ax )sin( bx ) dx 0

∞∞ 1 −−λλ =∫∫exxcos ( a +− b ) xdx e cos ( a − b ) xdx 2 00

1 λλ = − . 2λλ2++ ()ab 22 +− () ab 2

Integrating with respect to λ and evaluating the constant of integration with λ →∞ we find

∞ sin(ax )sin( bx ) 1λ 22++ (a b ) (2.24) J()λ = e−λx dx = ln . ∫ 22 0 x 4λ +− ()ab

This is entry 3.947(1) in [6].

Example 2.11

Using the previous example we can evaluate also entry 3.947(2) in [6].

∞ sin(ax ) sin( bx ) G()λ = e−λx dx ∫ 2 0 x

where again ab>>0 . We have from above

−1λλ2222 ++ (ab )1 +− (ab ) GJ′(λλ )=−= ( ) ln =ln 4λλ2222+− (ab )4 ++ (ab ) and integrating by parts,

λλ22+−(ab )1λ 2 λ 2 Gd(λλ )=−− ln 222222∫. 4λ++ ()4ab  λλ +− ()ab ++ () ab

With simple algebra we find

λλ22()ab+− 22 () ab −=− λλλλ2+−()ab 22 ++ () ab 22 ++ () ab 22 +− () ab 2

and the integration becomes easy. The result is

λλ22+−()ab ab − λ ab+ λ πb G(λ )=+−+ ln arctan arctan 4λ 22++ ()ab 2 ab− 2ab+ 2

(the constant of integration is found by setting λ →∞).

This answer is simpler than the one given in [6]. With λ = 0 we prove also 3.741(3) from [6]

∞ sin(ax ) sin( bx ) π b dx=( a ≥> b 0) . ∫ 2 0 x 2

Example 2.12

Similar to (2.24) is the integral

∞ sin(ax )cos( bx ) G()λ = ∫ e−λx dx 0 x

(this is 3.947(3) in [6]). Suppose ab>>0 . Then

∞ G′(λ )= −∫ e−λx sin( ax )cos( bx ) dx 0

∞∞ −1−−λλ =∫∫exxsin ( a ++ b ) xdx e sin ( a − b ) xdx 2 00

−+1ab ab − = + , 2λλ2++ ()ab 22 +− () ab 2

and after integration with respect to λ ,

π1  λλ G(λ )=−+ arctan arctan , 22ab+− ab

where the constant of integration π /2 is found by letting λ →∞.

Setting ba→ we find also

∞ sin(ax )cos( ax )πλ 1 ∫ e−λx dx = − arctan . 0 xa42 2

Using the identity 2sin(ax )cos( ax )= sin(2 ax ) this integral can be reduced to (1.4).

Example 2.13

∞ cos(ax )− cos( bx ) (2.25) F()λ = e−λx dx ∫ 2 0 x

(entry 3.948(3) in [6]). Differentiating we find

∞ cos(bx )−+ cos( ax ) 1 λ 22a F′()λ = e−λx dx = ln ∫ 22 0 xb2 λ +

in view of (2.18), as cos(bx )− cos( ax ) = cos( bx ) −+− 1 1 cos(ax ) . Integration by parts yields

λλ22+ a ba (2.26) F(λ )= ln +− ba arctan arctan 2 λ22+ b λλ

(again the constant of integration is found by setting λ →∞).

Various integrals with similar structure can be evaluated by this method or by reducing to those already evaluated here. For example, entry 3.948 (4) from [6]

∞ sin22 (ax )− sin ( bx ) A()λ = e−λx dx , ∫ 2 0 x can be reduced to (2.25) by using the identity 2sin2 xx= 1 − cos 2 . The results is

λλ22+ 4b 22 ab A(λ )= ln +− abarctan arctan . 44λ22+ a λλ

Now we shall evaluate several integrals involving logarithms of trigonometric functions.

Example 2.14

Consider the integral

π 2 (2.27) Jd(α )=∫ ln( α22 − cos θθ ) 0

for α >1. Differentiation with respect to α yields

π 2 dθ J′()αα= 2 ∫ 22 0 αθ− cos

and then the substitution x = tanθ turns this into

∞ ∞ dx 2 απx J′(αα )= 2 = arctan = . ∫ αα2−+ 22 2 22 0 1 x α−1 αα −− 110

Therefore,

π 2 Jd(α )=∫ ln( α22 − cos θθπ ) = ln( α + α2 −+ 1) C . 0

In order to evaluate the constant of integration we write this equation in the form (factoring out α 2 in the left hand side and α in the right hand side)

π 2 cos2 θ 1 παln+ ln 1 −dCθπαπ =ln + ln 1 +− 1 +. ∫ αα22 0  

Removing παln from both sides and setting α →∞ we compute C = −π ln 2 . As a result, two integrals are evaluated. For the second one we set βα=1/ in (2.27)

π 2 αα+−2 1 (2.28) ∫ ln(α22−= cos θθπ )d ln (α > 1) 0 2

π 2 11+−β 2 (2.29) ∫ ln(1−β22 cos θθπ )d = ln (0≤≤β 1) . 0 2

In particular, with β =1 in (2.29) we obtain the important log-sine integral

π 2 π (2.30) ∫ ln(sinθθ )d = − ln 2 . 0 2

Example 2.15

In the same way we can prove that

π /2 11++α (2.31) ∫ ln(1+=α sin2 θθπ )d ln 0 2

for any α >−1. Calling this integral F()α and differentiating we find

π /2 sin2 θ Fd′()αθ= ∫ 2 . 0 1+αθ sin

Now we divide top and bottom of the integrand by cos2 θ and then use the substitution x = tanθ

∞ x2 F′()α = dx . ∫ 22 0 (1+xx )(1 ++ (1α ) )

Assuming for the moment that α ≠ 0 and using partial fraction we write

∞ 11 1 F′()α = − dx ∫ 22 αα0 1+xx 1 ++ (1 )

∞ 1 1 π1  πα 11+− =arctanxx − arctan( 1 +α )= 1 −= . αα    1+α 0 2211++α  αα

Simple algebra shows that

π F′()α = 2(1++ 1αα ) 1 + which is exactly the derivative of παln(1++ 1 ) . Thus FC(απ )= ln(1 +++ 1 α ) . Setting α = 0 we find C = −π ln 2 and (2.31) is proved.

Example 2.16

Let |α |1< . Now we prove the interesting integral, entry 4.397 (3) in [6]

π ln (1+αθ cos ) (2.32) Fd()α≡=∫ θπarcsin α. 0 cosθ

Assuming that the value of the integrand at θπ= /2 is α , the integrand becomes a continuous function on [0,π ]. Then

π 1 Fd′()αθ≡ ∫ 0 1+αθ cos

θ 12− t2 dt which is easily solved with the substitution tan = t , so that cosθθ= , d = . Thus 2 11++tt22

∞∞ dt 2 dt F′()α = 2 = ∫∫2 1++−αα (1 )t 1 + α1−α 2 001+ t 1+α

∞ 21−απ = arctan t = , 221+α 11−−αα0 and (2.32) follows since both sides in this equation are zeros for α = 0 .

Example 2.17

Let again |α |1< . Using the results from the previous example we can prove

π 11+−α 2 (2.33) ∫ ln(1+=α cos θθπ )d ln . 0 2

Indeed, let this integral be F()α . We have

ππ π cosθ 1 1+− αθ cos 1 π 1 d θ Fd′()αθ= ∫∫= d θ= − ∫ 001++αθ cos α1 αθ cos αααθ 01 + cos

ππ = − α αα1− 2

(for the moment α ≠ 0 ). Integration is easy:

−1 dα −− FC(απ )= ln α + =παln + ln( α12 + α −+1) ∫ −2 ( ) α −1

=παln (1 +− 12 ) +C .

Now we can drop the restriction α ≠ 0 . With α = 0 we find C = −π ln 2 and (2.33) is proved.

Example 2.18

The last example in this section is a very interesting integral. It can be found, for example, in the book [12] on p. 143.

π (2.34) F()α=−+∫ ln(12cos ααx2 ) dx 0

We first assume α ≠ 0 and α ≠ 1. Then

ππ −+2cosx 2αα 1 1−2 F′()α = dx =1 − dx ∫∫22 0012cos−+αααxx12cos −+ αα

π πα11− 2 = − dx . ∫ 2 αα0 12cos−+ αx α

x 12− t2 dt The last integral can be evaluated by setting as before tan = t , with cosx= , dx = . 2 11++tt22

Some simple work gives

∞ πα21+ Ft′(α )= − arctan αα1− α0

2π and we find from here that F′()α = 0 when |α |1< and F′()α = when |α |1> . α

2 Thus FC()α = 1 when |α |1< and FC(α )=2 + πα ln for |α |1> . Since F(0)= 0 (as the definition (2.30) shows) we have

F()α = 0 for all |α |1< .

2 Next, to determine the constant C2 when |α |1> we factor out α inside the logarithm in (2.34) and write

π 1 2cos x  1 F(α )= ln α2 − +1dx =πα ln 2 + F = παln 22 += 0 πα ln , ∫ 2   0 αα α

2 that is, C2 = 0 and F(α )= πα ln for |α |1> . This also extends to α = ±1 , i.e.

F(α )= πα ln 2 for |α |1≥ .

It is good to mention here that the evaluation of this integral for the case |α |1< can be done immediately by using the series representation for 0 ≤≤x π

∞ α k cos kx (2.35) ln(12cos−ααx +=−2 ) 2∑ . k =1 k

The case |α |1> can be reduced to this one by writing

11 11 −+=−+=+−+ααα22 α ln(1 2 cosxx ) ln22 2 cos 1 2ln | | ln 2 cos x 1 . αα αα

The integral can be written in a symmetric form with |βα ||≤ | (cf. entry 2.6.36(14) in [8])

π (2.36) ∫ ln(β22− 2 αβ cosx += α ) dx 2 π ln | α | . 0

3. Using differential equations

Example 3.1 Consider the integral

∞ 2 y( x )= ∫ e−t cos (2 xt ) dt . 0

Here

∞ 2 y′( x )= −∫ 2 te−t sin (2 xt ) dt , 0

and integration by parts leads to the separable differential equation

dy y′ =−=−2 x y or 2xy dx

with general solution

2 y() x= Ce− x .

π For x = 0 in the original integral we have y(0) = according to (2.20). Therefore, 2

∞ 22π y( x )= ∫ e−tx cos (2 xt ) dt= e− . 0 2

With a simple rescaling of the variable we can write this result as

∞ 22π ∫ e−ax cos( xt ) d x= e−ta/4 ( a > 0) . −∞ a

This last integral was used in the solution of Problem 1896 of the Mathematics Magazine (vol. 83, June 2013, 228-230).

Example 3.2

In this example we evaluate two interesting integrals (3.723, (2) and (3) in [6])

∞∞cos λλx xxsin F()λλ= dx ,and() G = dx , ∫∫22 22 00ax++ax

which can be viewed as Fourier cosine and sine transforms. We shall use a second order differential equation for F()λ . First we have

(3.1) FG′()λλ= − ()

We cannot differentiate further, because G′()λ is divergent. Instead, we shall use a special trick, adding to both sides of (3.1) the number

π ∞ sin x = ∫ dx 2 0 x

(see (1.5)). After a simple calculation

πλ∞ sin x F′()λ += a2 dx . ∫ 22 20 xa ()+ x

Differentiating again we come to the second order differential equation

F′′ = aF2

with general solution

F()λ = Aeaaλλ + Be− ,

where AB, are arbitrary constants. Suppose a > 0 and λ ≥ 0 . Then A = 0 , because F()λ is a bounded function when λ →∞. To find B we set λ = 0 and use the fact that

∞ ∞ dx 1 x π BF=(0) = = arctan = . ∫ 22 0 ax+ a a0 2 a

Finally,

∞ cos λπx (3.2) F()λ = dx= e−aλ ∫ 22 0 ax+ 2 a

and from (3.1) we find also

∞ xxsin λπ (3.3) G()λ = dx= e−aλ . ∫ 22 0 ax+ 2

This result can be used to evaluate some similar integrals. Integrating (3.2) with respect to λ and adjusting the constant of integration we find entry 3.725 (1) [6]

∞ sin λπx dx=(1 − e−aλ ) . ∫ 22 2 0 xa(+ x )2 a

Differentiating this integral with respect to a we prove also entry 3.735

∞ sin λx π λπ dx=(1 −− e−−aaλλ ) e . ∫ 2 22 4 3 0 xa()2+ x a 4a

Example 3.3

We shall evaluate two Laplace integrals. For s > 0 and a > 0 consider

∞∞e−−st te st F() s= dt ,and G () s= dt . ∫∫22 22 00at++at

Differentiating twice the first one we find

(3.4) F′() s=−=− Gs (), F′′ G ′ () s

and at the same time

∞∞te2−st ()−+ a2 a 22 + t e−st 1 −=G′() s dt = dt=−+ a2 F() s ∫∫22 22 00at++ at s

which leads to the second order differential equation

1 F′′ += aF2 . s

This equation can be solved by variation of parameters. The solution is

1 F() s= [ci()sin()si()cos()] as as− as as a

involving the special sine and cosine integrals

∞ sin tt−π x sin si(x ) =−=+∫∫ dt dt , x tt2 0

∞ cost ci(x ) = −∫ dt . x t

The choice of integral limits here is dictated by the initial conditions FG()∞= () ∞= 0 .

From (3.4) we find also

G() s=−− ci()cos()si()sin() as as as as .

The integral Fs() can be used to give an interesting extension of the integral (1.5). For any ab,0> we compute

∞ ∞ ∞ ∞∞ sin(ax ) −−+−   ∫dx= ∫sin( ax )  ∫ et() x b dt  dx= ∫∫ ext sin( ax ) dx  ebt dt 00xb+ 0 00 

∞ e−bt dt =a = aF()ci()sin()si()cos() b = ab ab− ab ab . ∫ 22 0 ta+

This is entry 3.772 (1) from [6]. Taking limits of both sides when b → 0 yields (1.5). In the same way we prove entry 3.722(3)

∞ cos (ax ) ∫ dx=−−ci()cos()si()sin() ab ab ab ab . 0 xb+

Example 3.4

We shall evaluate Hecke’s integral

∞ α dx H (αα )=∫ exp −−x dx,> 0 0 x x by using a differential equation (cf. [7]). Differentiation yields

∞ α dx H′(α )=−∫ exp −−x dx 0 x xx and then the substitution xt= α / leads to the separable differential equation

−−1 dH dα HH′()αα= (),i.e. = ααH with solution

HM(αα )=−> exp( 2 ), M 0 a constant .

Setting here α = 0 and using the fact that H (0)=Γ= (1/ 2) π we find

(3.5) H (απ )= exp( − 2 α) .

Example 3.5

Consider the two integrals ([5], p.731)

∞ α 2 ∞ α 2 U (α )= exp( −x2 )cos dx V (α )= exp( −x2 )sin dx (3.6) ∫ 2 and ∫ 2 . 0 x 0 x

We differentiate U ()α and set yx= α / to find

∞ αα220 − α  U′(α )=−− 2 exp(x22 )sin dx =2 exp sin(y ) dy ∫∫22 2  0 xx∞  y 

∞ −α 2 = −2 exp sin(y2 ) dy ∫ 2 . 0 y

A second differentiation gives

∞∞−−αα22  2  α U′′(α )=−= 2 exp sin(y22 ) dy 4 exp(− x )sin dx ∫∫22  2 , 00yy   x that is

UV′′()αα= 4().

In the same way we compute VU′′()αα= − 4(). We define now the complex function W()αα= U () + iV () α. This function satisfies the second order differential equation.

WW′′ = −4i

2 with characteristic equation ri+=40 and roots ri1 =−+22 and ri2 =22 − . From these roots we construct the general solution to the differential equation

W()ααα= Ar exp(12 ) + Br exp( ) with parameters A and B . Explicitly,

WA(α )=− exp( 2 αα )(cos 2 ++i sin 2 α ) B exp( 2 αα )(cos 2 −i sin 2 α ) .

At this point we conclude that B = 0 , since W ()α is a bounded function. Setting α = 0 we find WA(0) = . At the same time by the definition (3.6) of the above integrals

∞ π WU(0)==−= (0)∫ exp(x2 ) dx 0 2 and

π Wi(α )=−+ exp( αα 2 )(cos( 2) sin( α 2)) . 2

From here, comparing real and imaginary parts we conclude that

π π (3.7) U (α )= exp( − αα 2 )cos( 2) , V ()α= exp( − αα 2)sin( 2) . 2 2

4. Advanced techniques

In certain cases we can use the Leibniz Integral Rule [5], [9]:

ψα ψα dd() () ∫∫f(,)α x dx=+− f (,)α x dx f (, αψ ())() α ψ′′ αf (,())() αϕα ϕ α , ddααϕα() ϕα() where fx(α , ), ϕα ( ), ψα ( ) are appropriate functions.

Example 4.1

We shall evaluate the integral

1 arctan x dx ∫ 2 0 1− x by using the function

1 arctan(α x ) J()α = dx ∫ 2 ϕα() 1− x where α >1 and

1 α 2 −1 1 ϕα =−= ′ () 1 2 with ϕα()= . αα αα22−1

Applying the Leibniz rule we find

1 x arctanα 2 − 1 J′()α = dx − . ∫ 22 2 2 ϕα()(1+−αxx ) 1 αα −1

Let us call this integral A()α . We shall evaluate it by the substitution 1−=x22 uu ,0 > :

1 1 x α du A()α = dx = ∫∫22 2 2 22 ϕα()(1+−α xx ) 1 0 αα+−1 u

1 11αα22++u α 111α++ = ln = ln . αα22+ α +− α αα22+ α +− 21 1u 0 21 11

This function is easy to integrate, as

d α 2 ++11 − 2 ln = dα α22+−11 αα + 1 and therefore, one antiderivative is

2 −1α 2 ++ 11 Ad(αα )=  ln . ∫ 2 8 α +−11

We also have

d 1 arctanα 2 −= 1 dα αα2 −1 and therefore,

arctanα 2 − 1 1 2 dαα= arctan2 − 1 . ∫ 2 ( ) αα−1 2

Now we can integrate J′()α to obtain

2 −1α 2 ++ 11 1 2 JC(αα )= ln −arctan2 −+ 1 . 2 ( ) 82α +−11

π 2 Using the limit limJ (α )= 0 we find C = . Finally, α →∞ 8

2 1 arctan()ααx − 122 ++ 11 1 2 π (4.1) dx=ln −arctanα 2 −+ 1 ∫ 22( ) ϕα() 1−x 828α +−11

Setting here α →1 we find after simple computation

1 arctan x π 2 1 2 (4.2) dx=−+ln (1 2 ) . ∫ 2 ( ) 0 1− x 82

With α = 2 in (4.1) we find also

2 1 arctan(2)x −+ 1 51 1 2 π 2 dx=ln −+arctan 3 . ∫ 2 ( ) 3/2 1− x 82851−

Remark. Integrating by parts in (4.2) we find

11arctan xxπ 2 arcsin dx= − dx ∫∫2 2 001− x 81+ x and therefore,

1 arcsinx 1 2 dx=ln (1 + 2 ) . ∫ 2 ( ) 0 12+ x

Using the identity π 1 arctanx = − arctan 2 x for x > 0 , we can write

1 112 arctan arctan x π dx= − x dx. ∫∫22 0011−−xx4

In the second integral we make the substitution xt=1/ to find also

∞ arctant π 2 1 2 (4.3) dt =++ln (1 2 ) . ∫ 2 ( ) 1 tt−1 82

This integral was evaluated in [4] independently of (4.2) by using the same method.

Example 4.2

We shall evaluate in explicit form the function

∞ ln(ααxx+−22 1) F()α = dx ∫ 2 1/α xx(1+ )

where α > 0 . Differentiating by the Leibniz rule we compute

∞ dx F′()α = ∫ 2 22 1/α (1+−xx )α 1

(notice that the function ln(ααxx+−22 1) becomes zero for x =1/α ).

To solve the integral we first write it in the form

−1 ∞ dx−2 F′()α = ∫ −−2 22 2 1/α (xx+− 1) α

and then we make the substitution α 2−=x− 22 tt,0 > to get

α dt 1 1++αα2 F′()α = = ln . ∫ 22 22 0 1+−α t 21+α 1 +− αα

This function is easy to integrate, as

d 1++αα2 2 ln = . dα 11+−αα22 + α

Thus we find

111++αα2 2 F(α )= ln 2=ln( 1 ++−αα22 ) ln( 1 +− αα ) 2 ( ) 881+−αα

(the constant of integration is zero, since F()α → 0 for α → 0) . Simplifying this we get

1 F(α )= ln22 ( 1 ++ αα ) 2

1 since ln( 1+−=αα22 ) ln =−++ln( 1αα ) . 1++αα2

Therefore, for any α > 0 ,

∞ ln(ααxx+−22 1) 1 (4.4) dx =ln22 ( 1 ++αα ) . ∫ 2 1/α xx(1+ ) 2

In particular, for α =1,

∞ ln(xx+−2 1) 1 dx =ln2 ( 2 + 1) (4.5) ∫ 2 . 1 xx(1+ ) 2

For α =1/2 in (4.4) we find

∞ ln(xx+−2 4) 5 1 5+ 1 dx =ln 2ln + ln 2 (4.6) ∫ 2 . 2 xx(1+ ) 2 2 2

Remark. Note that the similar integral

∞ ln(xx+−2 1) (4.7) dx ∫ 2 1 xx−1 cannot be evaluated this way. The value of this integral is 2G , where

∞ (− 1)n 1 1 = =−++ G ∑ 21 22 ... , n=0 (2n + 1) 3 5

is Catalan’s constant. The substitution xt= cosh with ln(xx+2 −= 1) t turns (4,7) into

∞ t ∫ dt= 2 G 0 cosh t

which is a well-known result.

5. Some theorems

Theorem A Suppose the function fx(,)α is defined and continuous on the rectangle

[,ab ]× [, cd ] together with its partial derivative fα (,)α x . Then

d dd ∫∫f(,)αα x dx= fα (,) x d x . dα cc

In order to apply this theorem in the case of improper integrals we have to require uniform convergence of the integral with respect to the variable α . A simple sufficient condition for uniform convergence is the following theorem

Theorem B Suppose fx(,)α is continuous on [ab , ]×∞ [0, ) and gx() is integrable on [0,∞ ) . If

|f (α , x )|≤ gx ( )

for all ab≤≤α and all x ≥ 0 , then the integral

∞ ∫ f(,)α x dx 0

is uniformly convergent on [,]ab.

Theorem C Suppose the function fx(,)α is continuous on [ab , ]×∞ [0, ) together with its

partial derivative fxα (,)α . In this case

∞∞ d ∫∫f(,)αα x dx= fα (,) x d x dα 00

when the first integral is convergent and the second is uniformly convergent on [,]ab.

The case of improper integrals on finite intervals is treated in the same way. For details and proofs we refer to [5], [7], [9], and [12]. The book [5] presents the Leibniz rule in full detail.

References

[1] Khristo N. Boyadzhiev, Some integrals related to the Basel problem, SCIENTIA. Series A: Mathematical Sciences, 26 (2015), 1-13. Also arXiv:1611.03571 [math.NT]

[2] Khristo N. Boyadzhiev, On a series of Furdui and Qin and some related integrals, 2012, arXiv:1203.4618v3 [math.NT]

[3] Khristo Boyadzhiev, Hans Kappus, Solution to problem E 3140, Amer. Math. Monthly, 95 (1), (1988), 57-59.

[4] Hongwey Chen, Parametric differentiation and integration, Internat. J. Math. Ed. Sci. Tech. 40 (4) (2009), 559-579.

[5] G. M. Fikhtengol’ts, A Course of Differential and Integral Calculus (Russian), Vol. 2, Nauka, Moscow, 1966.

Abbreviated version in English: The Fundamentals of Mathematical analysis, Vol. 2, Pergamon Press, 1965.

[6] I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series, and Products, Academic Press, 1980.

[7] Omar Hijab, Introduction to Calculus and Classical Analysis, Springer, 1997.

[8] A. P. Prudnikov, Yu. A. Brychkov, O. I. Marichev, Integrals and Series, vol.1 Elementary Functions, Gordon and Breach 1986.

[9] Ioannis Markos Roussos, Improper Riemann Integrals, Chapman and Hall/CRC, 2014.

[10] Joseph Wiener, Differentiation with respect to a parameter, College Mathematics Journal, 32, No. 3.(2001), pp. 180-184.

[11] 66th Annual William Lowell Putnam Mathematical Competition, Math. Magazine, 79 (2006), 76-79.

[12] Frederick. S. Woods, Advanced Calculus: A Course Arranged with Special Reference to the Needs of Students of Applied Mathematics. New Edition, Ginn and Co., Boston, MA, 1934.

[13] Aurel J. Zajta, Sudhir K. Goel, Parametric integration techniques, Math. Magazine, 62(5) (1989), 318-322.