Methods on finding closed form of $\int_0^\infty x^{s-1}e^{-a(x+\frac{1}{x})}dx$

Question

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.

Answer 1

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

$$K_{s}(2a)=\int_0^\infty e^{-2a\cosh(t)}\,\cosh(st)\,dt\tag1$$

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

$$\begin{align} K_s(2a)&=\frac12\int_1^\infty e^{-a(x+1/x)}\left(x^{s-1}+1/x^{s+1}\right)\,dx\\\\ &=\frac12\int_1^\infty e^{-a(x+1/x)}x^{s-1}\,dx+\frac12\int_1^\infty e^{-a(x+1/x)}\frac1{x^{s+1}}\,dx\tag2 \end{align}$$

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

$$K_s(2a)=\frac12\int_0^\infty x^{s-1}e^{-a(x+1/x)}\,dx$$

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

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

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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 $b>0$

$$J(s,a,b)=\frac1b\int_0^\infty x^{s/b-1} e^{-a(x+1/x)}\,dx=\frac2b K_{s/b}(2a)$$

And we are done!

Answer 2

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.