Divergent Integrals of Certain Analytical Functions in the Sense of Zeta Regularization

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Suppose the following integral of function f(x) diverges ∞ I = f(x)dx (1) Z0 Also assume that the function has a convergent Maclaurin series with infinity radius of convergence, i.e., ∞ xk f(x)= f (k)(0) x R (2) k! ∀ ∈ k X=0 Then we seek to assign a bounded meaningful solution to the divergent integral (1) using a Zeta regularization technique. Note that if function f(z) is equal to its Maclaurin series for all z in the complex plane, then it is called entire. Examples of entire functions include polynomials, exponential function, and the trigonometric functions sine and cosine, while square root and the fractional functions are not entire. It is worthwhile mentioning that the necessary and sufficient for function f(z) in terms of its complex variable z = x+iy to be analytic are that: i) ∂u/∂x, ∂u/∂y, ∂v/∂x, ∂v/∂y are continuous and ii) the aforementioned four partial derivatives of the real part u(x, y) and imaginary part v(x, y) of the function satisfies the Cauchy-Riemann equations ∂u ∂v ∂u ∂v = and = (3) ∂x ∂y ∂y −∂x Replacing the function with its Maclaurin series (2) in the integral (1) and then switching the integral and sum, we can rewrite the divergent integral as follows ∞ f (k)(0) ∞ I = xk dx (4) k! k 0 X=0 Z The above improper integral can be equivalently written as the following infinite series by splitting the range of integration limits into successive integer numbers ∞ ∞ ∞ n 1 xk dx = xkdx = nk+1 (n 1)k+1 (5) − k +1 − − 0 n=1 n 1 n=1 Z X Z X Using (5) in (4) yields ∞ ∞ f (k)(0) I = nk+1 (n 1)k+1 (6) (k + 1)! − − r=0 n=1 X X According to the binomial polynomial expansion of (n 1)k+1, we have − k +1 k +1 (n 1)k+1 = ( 1)k+1−pnp, (7a) − p − p=0 X where the binomial coefficient is the defined as m m! = (7b) l l!(m l)! − 2 Upon substitution of (7a) in (6), we arrive at

∞ ∞ k f (k)(0) k +1 I = ( 1)r−pnp (8) (k + 1)! p − r=0 n=1 p=0 X X X ∞ k ∞ f (k)(0) k +1 = ( 1)k−p np (9) (k + 1)! p − k p=0 n=1 X=0 X X ∞ k f (k)(0) k +1 = ( 1)k−pζ( p) (10) (k + 1)! p − − k p=0 X=0 X For positive integer p 0, the Zeta function is related to Bernoulli numbers by ≥ B ζ( p)=( 1)p p+1 (11) − − p +1 Thus, we have ∞ k (k) k ( 1) f (0) k +1 Bp I = − +1 (12) (k + 1)! p p +1 k p=0 X=0 X Moreover, from deﬁnition (7b), one can verify that the successive binomial coeﬃcients hold the following useful identity k +1 p +1 k +2 = (13) p k +2 p +1 Thus we have ∞ k ( 1)kf (k)(0) k +2 I = − Bp (k + 2)! p +1 +1 k p=0 X=0 X ∞ k ( 1)kf (k)(0) +1 k +2 = − Bp (k + 2)! p k p=1 X=0 X ∞ k ( 1)kf (k)(0) +1 k +2 k +2 = − Bp B (k + 2)! p − 0 0 k p=0 ! X=0 X ∞ k ( 1)kf (k)(0) +1 k +2 = − Bp 1 (14) (k + 2)! p − k p=0 ! X=0 X On the other hand, the Bernoulli numbers satisfy the following property

m− k 1 m +1 k +2 B = B =0 (15) p p p p p=0 p=0 X X Finally using identity (15) in (14), we arrive at the explicit expression of the divergent integral based on Zeta regularization

∞ ∞ ( 1)k+1 f(x) dx = − f (k)(0) (16) (k + 2)! 0 k Z X=0 3 Invoking the ratio test, one can conclude that that the above inﬁnite series converge if

(−1)k+2f (k+1)(0) (k+3)! lim k k < 1 (17) k→∞ (−1) +1f ( )(0) (k+2)!

In other words, the convergence of the above series is

f (k+1)(0) lim | | < 1 (18) k→∞ (k + 3) f (k)(0) | | Equation (16) can be used to derive Zeta regularization of many divergent integrals. For instance, if the analytical function is selected to be a polynomial

f(x)= xm (19)

Then, m! if k = m f (k)(0) = (20) 0 otherwise Applying (16) to (20) yields

∞ ( 1)m+1 xm dx =0+0+ + − m!+0+0+ , ··· (m + 2)! ··· Z0 and thus ∞ ( 1)m+1 xm dx = − , (m + 1)(m + 2) Z0 which is consistent with the previous result in [3].

3 Examples of Divergent integrals

Consider the trigonometric function

f(x) = sin(x) (21)

whose derivatives are f (2k+1)(0) = ( 1)2k+1. Thus, according to (16) we have − ∞ ∞ ( 1)k+1 sin(x) dx = − (22) (2k + 3)! 0 k Z X=0 By inspection one can show that the right-hand side of the above equation is equal to sin(1) 1. −

4 Examples of Zeta regularization of divergent integrals of several transcendental functions are given below:

∞ ∞ ( 1)k+1 sin(x)dx = − = sin(1) 1 (23a) (2k + 3)! − Z0 k=0 X∞ ∞ ( 1)2k+1 cos(x)dx = − = cos(1) 1 (23b) (2k + 2)! − Z0 k=0 X∞ ∞ 1 sinh(x)dx = = sinh(1) 1 (23c) (2k + 3)! − Z0 k=0 X∞ ∞ ( 1)2k+1 cosh(x)dx = − =1 cosh(1) (23d) (2k + 2)! − Z0 k=0 ∞ X ∞ ( 1)k+1 1 exdx = − = (23e) (k + 2)! −e Z0 k=0 X ∞ ∞ 1 1 log(1 + x)dx = = (23f) k(k + 1)(k + 2) 4 Z0 k=0 ∞ X ∞ k+1 2 ( 1) 1 1 √π ex dx = − = erf(1) (23g) 2(k + 2)(k + 1)k! 2 − 2e − e Z0 k=0 X ∞ ∞ 2 ( 1)k 3 1 1 erf(x)dx = − = erf(1) + (23h) √π 2(2k + 1)(k + 1)(2k + 3)k! 2 2e√π − √π Z0 k=0 ∞X ∞ k+1 − 2 2 ( 1) √π 1 e e x dx = − = erf(1) + − (23i) √π 2(2k + 1)(k + 1)! − 2 2e Z0 k=0 X∞ ∞ 1 x sin(x)dx = = 2 + sinh(1) 2cosh(1) (23j) 2(k + 2)(2k +3)(1+2k)! − 0 k Z X=0 4 Conclusions

We extended the zeta function regularization technique in order to assign finite values to divergent integral of a class of analytical functions which satisfy a convergence condition. Using the binomial expansion and Maclaurin series, it has been shown that such divergent integrals can be expressed by infinite series in terms of the Riemann zeta function. Subsequently, it has been shown that the divergent integral can be written into the following infinite series k ∞ ∞ (−1) +1 (k) 0 f(t)dt = k=0 (k+2)! f (0), which becomes convergent under certain condition. Calcula- tion of several common divergent integrals using this formula has been presented. R P References

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