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?
 A: 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 $B(l+1,\frac12-l)$.
A: 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.
