Ytukyg

I was playing around with Mellin Transforms and encountered this interesting integral$$I(s,a)=int_0^infty x^{s-1}e^{-a(x+frac{1}{x})}dx$$

Since $f(x)=e^{-a(x+frac{1}{x})}=f(1/x)$, I found that the integral is even with respect to $s$
$$I(s,a)=I(-s,a)$$

Are there any methods for evaluating this integral? I thought about using Ramanujan's Master Theorem but I cannot get $f(x)$ in a form where I can apply it. Furthermore, when I used Wolfram to evaluate this integral at $a=1$, it comes back with a form of the Modified Bessel Function. Maybe the closed form could be obtained in terms of Bessel Functions of some sort?

Also, it would be interesting to see if a closed form can be found for a more general integral $$J(s,a,b)=int_0^{infty} x^{s-1}e^{-a(x^{b}+x^{-b})}dx$$
that satisfies $J(s,a,b)=J(-s,a,b)$. I would be grateful for anyhelp as I am particularly intrigued by these integrals' properties.

The Second Kind modified Bessel function of order $s$ and argument $2a$ can be represented by the integral (See This).

$$K_{s}(2a)=int_0^infty e^{-2acosh(t)},cosh(st),dttag1$$

Enforcing the substitution $x=e^t$ in $(1)$ reveals

$$begin{align}
K_s(2a)&=frac12int_1^infty e^{-a(x+1/x)}left(x^{s-1}+1/x^{s+1}right),dx\\
&=frac12int_1^infty e^{-a(x+1/x)}x^{s-1},dx+frac12int_1^infty e^{-a(x+1/x)}frac1{x^{s+1}},dxtag2
end{align}$$

Finally, enforce the substitution $xmapsto 1/x$ in the second integral in the right-hand side of $(2)$ yields

$$K_s(2a)=frac12int_0^infty x^{s-1}e^{-a(x+1/x)},dx$$

from which we conclude that the integral of interest $I(s,a)$ is

$$I(s,a)=2K_s(2a)$$

EDIT: Motivated by a comment from @Fabian

Suppose now that $J(s,a,b)=int_0^infty u^{s-1} e^{-a(u^b+1/u^b)},du$. Letting $u=x^{1/b}$, we find that for $b>0$

$$J(s,a,b)=frac1bint_0^infty x^{s/b-1} e^{-a(x+1/x)},dx=frac2b K_{s/b}(2a)$$

And we are done!

Using integration by parts, we may establish a sort of recurrence:
$$begin{align}
I(s,a)&=int_0^infty x^{s-1}e^{-ax-a/x}dx\
&=bigg[frac{x^se^{-ax-a/x}}{s}bigg]_0^infty+frac{a}{s}int_0^infty x^{s}e^{-ax-a/x}bigg(1-frac{1}{x^2}bigg)dx\
&=frac{a}{s}int_0^infty x^{s}e^{-ax-a/x}dx-frac{a}{s}int_0^infty x^{s-2}e^{-ax-a/x}dx\
&=frac{a}{s}I(s+1,a)+frac{a}{s}I(s-1,a)
end{align}$$
Wolfram evaluates $I(1,a)$ as $2K_1(2a)$ and $I(2,a)$ as $2K_2(2a)$, where $K_n(x)$ is the modified Bessel function of the second kind. Thus, we have a recurrence
$$I(s+1,a)=frac{s}{a}I(s,a)-I(s-1,a)$$
with initial values $I(1,a)=2K_1(2a)$ and $I(2,a)=2K_2(2a)$. This yields a closed-form (for positive integer $s$) as
$$I(s,a)=2(s-1)!sum_{i=1}^{s-1} frac{K(2a)}{a^i i!}$$
where the subscript for $K$ alternates between $2,1,2,1$ etc.