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?



Improper integral $\int^\infty_0 e^{-a x^2} \cosh (b\sqrt{1+x^2})$
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 $B(l+1,\frac12-l)$.


