# Improper integral $\int^\infty_0 e^{-a x^2} \cosh (b\sqrt{1+x^2})$

In my research, I ran into following types of improper integral

$\int^\infty_0 e^{-a x^2} \cosh (b\sqrt{1+x^2})$

with real parameters $a>0,b>0$.

Mathematica cannot evaluate them. It also seems that a definite integral of sort

$\int \cos (\sqrt{1+x^2})$, $\int \cosh (\sqrt{1+x^2})$ ,… etc

cannot be evaluated in mathematica, and i couldn't find them in the tables yet.

Could anyone help me how to solve these integrals?

• In the case $a=b=1$ I calculated the integral to 50 digits using Maple and gave the result to the Inverse Symbolic Calculator at oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html; it was not recognized. Given that neither Maple, Mathematica nor the ISC knows an answer in closed form, I doubt that one exists. – Neil Strickland May 20 '14 at 15:11
• I would try using a Mellin transform approach. You can evaluate the Mellin transform of both functions in your integrand (I was able to using Mathematica, taking 1+x^2 -> u^2) in cosh term), then use the convolution theorem to write the integral over a vertical line in the complex plane. The trick then is to close the contour and evaluate the integral as the sum of residues from the poles of Gamma functions. There are many references to this, a good paper is "Evaluation of integrals and the mellin transform," by Prudnikov, et al. Mathematica uses this technique internally. – Tom Dickens May 20 '14 at 20:07
• By the way, Liouville's theorem and the Risch algorithm are about closed-form antiderivatives, not definite integrals. – Tom Dickens May 20 '14 at 20:08
• @TomDickens: Correct! Thanks for pointing that out! – Lucian May 20 '14 at 21:03
• @TomDickens: Thanks for the suggestion. I will look into the paper, and give it a try. If mathematica uses this method internally but can't evaluate the integral, do I have a chance to get to a closed form by doing it by myself? – user113103 May 21 '14 at 15:55

Let us give only an expansion in $a,b$. Calling $I(a,b)$ the integral, we get easily $$I(a,b)=\sum_{k\ge 0}\frac{b^{2k}}{(2k)!}\underbrace{e^{a}\int_0^{+\infty} e^{-a (x^2+1)}(1+x^2)^k dx}_{J_k(a)}.$$ We have $J_k(a)=e^{a}(-\frac{d}{da})^k\bigl(J_0(a)\bigr)=\sqrt π e^{a}(-\frac{d}{da})^k\bigl(e^{-a}a^{-1/2}\bigr).$ Defining for $\alpha\notin-\mathbb N^*$, $$\chi_+^\alpha(x)=\frac{x^\alpha}{\Gamma(\alpha +1)}\mathbf 1_{\mathbb R_+}(x), \quad \text{ we see that }\quad \frac{d\chi_+^\alpha}{dx}=\chi_+^{\alpha-1},$$ we get, since $\Gamma(1/2)=\sqrtπ$ \begin{align} I(a,b)&=π\sum_{k\ge l\ge 0}\frac{(-1)^kb^{2k}}{(2k)!} \chi_+^{-\frac{1}{2}-l}(a)(-1)^{k-l}\frac{k!}{l!(k-l)!}\\&= 2\sqrtπ\sum_{k\ge l\ge 0}\frac{(-1)^kb^{2k}}{(2k)!} \frac{a^{-\frac 12-l}}{\Gamma(\frac12-l)}(-1)^{k-l}\frac{k!\Gamma(3/2)}{\Gamma(l+1)(k-l)!} \\ &=2\sqrtπ\sum_{k\ge l\ge 0}\frac{b^{2k}}{(2k)!} \frac{a^{-\frac 12-l}}{B(l+1,\frac12-l)}(-1)^{l}\frac{k!}{(k-l)!}, \end{align} where $B$ stands for the Beta function. Well, it is not so friendly, but slightly more "explicit" than the initial formula. A little more effort will allow the reader to compute explicitly $B(l+1,\frac12-l)$.

• Thank you for the answer. Actually, it might be that your answer is explicit enough for my purpose, which is to do inverse laplace transformation of I(a,b). I don't follow the derivation completely, however. What is $1_{R_+}(x)$? Also, how $J_k(a)$ becomes a sum for $l$ is not clear to me. Could you explain a little bit more on this? – user113103 May 21 '14 at 16:05
• As @TomDickens suggested in the comment below, Mathematica could evaluate $J_k(a)$. Mathematica gives $J_k(a)= \frac{1}{2} e^{-a} \left(a^{-k-\frac{1}{2}} \Gamma \left(k+\frac{1}{2}\right) \, _1F_1\left(-k;\frac{1}{2}-k;a\right)+\frac{\sqrt{\pi } \Gamma \left(-k-\frac{1}{2}\right) \, _1F_1\left(\frac{1}{2};k+\frac{3}{2};a\right)}{\Gamma (-k)}\right)$ where $_1F_1$ refers to confluent hypergeometric function of the first kind. – user113103 May 21 '14 at 16:14
• @user113103 The indicatrix function of $\mathbb R_+$ is the standard Heaviside function, which is 1 on the positive half-line and 0 on the negative half-line. $J_k(a)$ appears essentially as the $k$-th derivative of a product: just apply Leibniz formula. – Bazin May 21 '14 at 19:05

Although there is no closed form in terms of elementary functions, $$\int^\infty_0 e^{-a x^2} \cosh (bx)dx=\frac12\exp\bigg(\frac{b^2}{4a}\bigg)\sqrt\frac\pi a$$ should make for a decent lower limit. Though, depending on the ratio between a and b, its value

can be up to several $/$ many times smaller than that of the original. In any case, the asymptotic

approximation can be significantly improved, by adding a second approximating function for the

integrand, for small values of x, since the one mentioned above works better for larger values of

the variable, as $x\to\infty$. Hope this helps.

• Note that the integral $J_k(a)$ in the above answer is a confluent hypergeometric function. Mathematica can do it or you can find it in a handbook I imagine. Sorry I don't have it in front of me right now. That will at least give you a compact form. – Tom Dickens May 21 '14 at 2:17
• @TomDickens: The Wikipedia articles on the Bessel function of the first kind and confluent hypergeometric series are quite comprehensive, as are those of NIST DLMF and Wolfram Math World. – Lucian May 21 '14 at 3:41