7.8 Improper Integrals Improper Integral Type I (on infinite intervals) a] Assume that f is continuous on [a, ). Then ∞ ∞ t f(x) dx := lim f(x) dx. Za t→∞ Za b] Assume that f is continuous on ( ,a]. Then −∞ a a f(x) dx := lim f(x) dx. Z−∞ t→−∞ Zt c] Assume that f is continuous on ( , ). Then −∞ ∞ ∞ a ∞ f(x) dx := f(x) dx + f(x) dx. Z−∞ Z−∞ Za (It can be shown that the value of the integral on ( , ) is independent of the choice of a, we can choose a = 0, for instance.) −∞ ∞ In parts a] or b] we say that the improper integral converges (or the improper integral is convergent) if the limit in the definition exists as a finite real number. In any other cases we say that the improper integral diverges. It can diverge “nicely” to or if the limit in the definition exists as or , but the improper integral can diverge∞ −∞ “due to oscillation” as well (roughly speaking).∞ −∞ In part c] we say that the improper integral converges (or the improper integral is convergent) if both integrals in the right-hand side converges. In any other cases we say that the improper integral diverges. Recall that + L = , + L = , + = , and ( )+( ) = . ∞ ∞ −∞ −∞ ∞ ∞ ∞ −∞ −∞ −∞ E.1. We show that ∞ 1 1 p dx = , p > 1 , Z1 x p 1 − and ∞ 1 p dx = , p 1 . Z1 x ∞ ≤ In other words ∞ 1 p dx Z1 x converges to 1/(p 1) for all p > 1, while it diverges to for all p 1. − ∞ ≤ Using the definition we have ∞ t t −p+1 t 1 1 −p x p dx = lim p dx = lim x dx = lim Z1 x t→∞ Z1 x t→∞ Z1 t→∞ p +1 − 1 t−p+1 1 1 1 = lim = 0 = t→∞ p +1 − p +1 − p +1 p 1 − −1 − − for all p > 1 as p +1 < 0 in the exponent of t tending to . Also, − ∞ ∞ t t −p+1 t 1 1 −p x p dx = lim p dx = lim x dx = lim Z1 x t→∞ Z1 x t→∞ Z1 t→∞ p +1 − 1 t−p+1 1 1 = lim = = t→∞ p +1 − p +1 ∞ − p +1 ∞ − − − for all p < 1 as p +1 > 0 in the exponent of t tending to . Finally for p = 1 we have − ∞ ∞ t t 1 1 −1 t p dx = lim dx = lim x dx = lim [ln x]1 Z1 x t→∞ Z1 x t→∞ Z1 t→∞ = lim (ln t ln1) = 0 = . t→∞ − ∞ − ∞ The above example will be quite useful later in this section when we detect convergence or divergence of improper integrals by using comparison theorems. ∞ 1 E.2. Show that p dx converges for all p > 1, while it diverges to for all p 1. Ze x(ln x) ∞ ≤ Hint: Use a substitution u = ln x to observe that ∞ 1 ∞ 1 p dx = p du, Ze x(ln x) Z1 u and recall the result of E.1. ∞ 1 E.3. Show that p dx converges for all p > 1, while it diverges to for Zee x(ln x)(lnln x) ∞ all p 1. Hint:≤ Use a substitution u = lnln x to observe that ∞ 1 ∞ 1 p dx = p du, Zee x(ln x)(ln ln x) Z1 u and recall the result of E.1. ∞ E.4. Does xe−x dx converge? If it does, compute its value. Z0 Using the definition we get ∞ t xe−x dx = lim xe−x dx. Z0 t→∞ Z0 The Integration by Parts (IP) formulas with f(x) = x, g′(x) = e−x, f ′(x) = 1 and g(x) = e−x gives − xe−x dx = x( e−x) 1( e−x) dx = xe−x e−x , Z − − Z − − − 2 hence t −x −x −x t −t −t t +1 xe dx =[ xe e ]0 = te e ( 1)=1 t , Z0 − − − − − − − e and taking the limit with an application of the L’Hospital’s Rule gives ∞ t −x −x t +1 t +1 xe dx = lim xe dx = lim 1 t =1 lim t Z0 t→∞ Z0 t→∞ − e − t→∞ e 1 1 = 1 lim =1 =1 0=1 . − t→∞ et − − ∞ ∞ E.5. Does e−|x| dx converge? If it does, compute its value. Z−∞ Using the definition in part c] we get ∞ 0 ∞ e−|x| dx = e−|x| dx + e−|x| dx =1+1=2 . Z−∞ Z−∞ Z0 Using the definition in parts a] and b] here we have ∞ ∞ t −|x| −x −x −x t e dx = e dx = lim e dx = lim [ e ]0 Z0 Z0 t→∞ Z0 t→∞ − = lim ( e−t ( e0))=1 0=1 , t→∞ − − − − and 0 0 0 −|x| x x x 0 e dx = e dx = lim e dx = lim [e ]t Z−∞ Z−∞ t→−∞ Zt t→−∞ = lim (e0 et)=1 0=1 , t→−∞ − − ∞ 4x E.6. Does 4 dx converge? If it does, compute its value. Z0 x +4 If we substitute u = x2/2 with du/dx = x, then we get 4x 4 1 dx = du = du = arctan u = arctan(x2/2)(+C) , Z x4 +4 Z 4u2 +4 Z u2 +1 hence ∞ t 4x 4x 2 t 4 dx = lim 4 dx = lim [arctan(x /2)]0 Z0 x +4 t→∞ Z0 x +4 t→∞ π π = lim (arctan(t2/2) arctan0) = 0 = . t→∞ − 2 − 2 3 ∞ E.7/a. Does sin xdx converge? If it does, compute its value. Z0 The definition gives ∞ t sin xdx := lim sin xdx Z0 t→∞ Z0 However if n is an even integer, then nπ nπ sin xdx =[ cos x]0 =( 1) ( 1)=0 , Z0 − − − − while if n is an odd integer, then nπ nπ sin xdx =[ cos x]0 = ( 1) ( 1)=1+1=2 . Z0 − − − − − Hence, by recalling the concept of limit of f(t) when t , we see that the limit does t → ∞ not exist as the function f(t) := sin xdx oscillates between 0 and 2 for arbitrarily large Z0 t values of t. So lim sin xdx does not exist (not even as an extended real number), so t→∞ Z0 ∞ we have to say that sin xdx diverges due to oscillation (roughly speaking). Z0 ∞ E.7/b. Show that sin x dx diverges to . Z0 | | ∞ Hint: This follows from the observation that t sin x dx Z0 | | is as large positive number as we wish for all sufficiently large values of t. ∞ ln x E.8. Does 2 dx converge? If it does, compute its value. Z1 x Recall that the Integration by Parts Formula gives that ln x x−1 ln x dx = (ln x)x−2 dx = (ln x) x−1( x−1) dx = + x−2 dx Z x2 Z 1 − Z − − x Z − ln x x−1 = + . − x 1 − We have ∞ ln x t ln x ln x x−1 t 2 dx = lim 2 dx = lim + Z1 x t→∞ Z1 x t→∞ − x 1 − 1 ln t 1 ln1 1 = lim t→∞ − t − t − − 1 − 1 ln t 1 1/t = 1+ lim − − =1+ lim − =1+0=1 , t→∞ t t→∞ 1 where at the end we used L’Hospital’s Rule to calculate an / type limit. ∞ ∞ 4 Improper Integral Type II (on finite intervals) a] Assume that f is continuous on (a,b]. Then b b f(x) dx := lim f(x) dx. + Za t→a Zt b] Assume that f is continuous on [a,b). Then b t f(x) dx := lim f(x) dx. Za t→b− Za c] Assume that f is continuous on [a,b] except at c [a,b]. Then ∈ b c b f(x) dx := f(x) dx + f(x) dx. Za Za Zc It can be shown that the value of the integral on [a,b] is independent of the choice of c. In parts a] or b] we say that the improper integral converges (or the improper integral is convergent) if the limit in the definition exists as a finite real number. In any other cases we say that the improper integral diverges. It can diverge “nicely” to or if the limit in the definition exists as or , but the improper integral can diverge∞ −∞ “due to oscillation” as well (roughly speaking).∞ −∞ In part c] we say that the improper integral converges (or the improper integral is convergent) if both integrals in the right-hand side converges. In any other cases we say that the improper integral diverges. Recall that + L = , + L = , + = , and ( )+( ) = . E.9.∞ We show∞ −∞ that −∞ ∞ ∞ ∞ −∞ −∞ −∞ 1 1 1 dx = , p < 1 , Z xp 1 p 0 − and 1 1 p dx = , p 1 . Z0 x ∞ ≥ In other words 1 1 p dx Z0 x converges to 1/(1 p) for all p < 1, while it diverges to for all p 1. − ∞ ≥ Using the definition we have 1 1 1 −p+1 1 1 1 −p x p dx = lim p dx = lim x dx = lim Z0 x t→0+ Z x t→0+ Z t→0+ p +1 t t − t 1 t−p+1 1 1 = lim = 0 = t→0+ p +1 − p +1 1 p − 1 p − −5 − − for all p < 1 as p +1 > 0 in the exponent of t tending to 0+. Also, − 1 1 1 1 1 1 x−p+1 dx = lim dx = lim x−p dx = lim Z xp t→0+ Z xp t→0+ Z t→0+ p +1 0 t t − t 1 t−p+1 1 = lim = ∞ = t→0+ p +1 − p +1 p +1 − p +1 ∞ − − − − for all p > 1 as p +1 < 0 in the exponent of t tending to 0+.