# Integral of combination of logarithm, power, and exponential, Modified Bessel.

I am trying to evaluate the following integral, with $$m\in \mathbb{N}$$, $$a, b \in \mathbb{R}$$, $$a>0, b>0$$.

$$\int_{0}^{\infty}\frac{(\log{x})^m}{x}\exp(-ax - b/x)\,dx.$$

I can find answers for related integrals, for example from "Table of integrals, series, and product" p.575: $$\int_0^{\infty}\exp{\left[\mu\left(\frac{x}{a} + \frac{a}{x}\right)\right]}\ln{x}\,\frac{dx}{x} = 2\ln{a}K_0(2\mu).$$ But I haven't been able to find anything with exponent $$m\geq2$$.

For $$m=1$$, the answer from wolfram alpha is $$\int_{0}^{\infty}\frac{\log{x}}{x}\exp(-ax - b/x)\,dx = \log{\left(\frac{b}{a}\right)}K_0(2\sqrt{ab}).$$ Where $$K_0(x)$$ is the modified Bessel function of the second kind. I am pretty sure that the connection to $$K_0(x)$$ comes from the following integral representation: $$K_{\alpha}(x) = \int_{0}^{\infty}\exp{(-x\cosh{t})}\cosh{\alpha t}\,dt.$$

The integrand vanishes at $$0$$ and $$\infty$$, so it is possible to use integration by parts to rewrite the integral in terms of similar integrals of the form:

$$f(k, m) = \int_{0}^{\infty}\frac{(\log{x})^{m'}}{x^{k'}}\exp(-ax - b/x)\,dx$$ with different values $$m',\,k'$$. as long as $$m'=1$$ Wolfram alpha can provide analytical solutions, even though they get messy. One possible route would be to recursively rewrite the integral in terms of similar integrals with lower $$m'$$-values, and eventually get down to the base case with $$m'=1$$. But I don't know if that is feasable.

(Note: To the question of what kind of answer I am looking for. I want a closed form solution of some kind. Expressing the integral in terms of the modified bessel function of the second kind seems like a good candidate. For different values of $$m$$ the solution will look different of course. It might not be possible to get a neat expression that works for all $$m$$? But at least get some sort of procedure to calculate.)

EDIT I think I have it now. The integral can be written as a Mellin transform. And multiplication by a factor $$(\log{x})^{m^{'}}$$ is the same as differentiating the transform $$m$$ times. After some quick and dirty, I got

$$\mathcal{M}\{\exp(-ax -b/x)\} = 2\left(\frac{b}{a}\right)^{s/2}K_{-s}(2\sqrt{ab})$$

So just differentiate that expression w.r.t. $$s$$, $$m$$ times, and then evaluate at $$s=0$$.

• $\log x^m=m\log x$ reduces the cases to $m=1$. Commented Jan 10, 2022 at 22:21
• by $\log{x}^m$ I mean $(\log{(x)})^m$. Thanks for your comment, I will edit to clarify Commented Jan 10, 2022 at 23:01
We have \begin{align*} &\int_0^{ + \infty } {\frac{{(\log x)^m }}{x}\exp ( - ax - b/x)dx} = \int_0^{ + \infty } {\left[ {\frac{{d^m }}{{d\nu ^m }}x^{\nu - 1} } \right]_{\nu = 0} \exp ( - ax - b/x)dx} \\ & = \left[ {\frac{{d^m }}{{d\nu ^m }}\int_0^{ + \infty } {x^{\nu - 1} \exp ( - ax - b/x)dx} } \right]_{\nu = 0} = 2\left[ {\frac{{d^m }}{{d\nu ^m }}\left( {\left( {\frac{b}{a}} \right)^{\nu /2} K_\nu (2\sqrt {ab} )} \right)} \right]_{\nu = 0} . \end{align*} The case $$m=0$$ is indeed expressible in terms of $$K_0$$ using the fact that $$\left[ {\frac{d}{{d\nu }}K_\nu (z)} \right]_{\nu = 0} = 0$$ (see below). The higher derivatives are more complicated objects. The first few of them are given here. An integral representation at $$\nu=0$$ is $$\left[\frac{{d^m }}{{d\nu ^m }}K_\nu(z)\right]_{\nu = 0} = \frac{1}{2}\int_{ - \infty }^{ + \infty } {x^m \exp ( - z\cosh x)dx} .$$ You can see that these will vanish for odd $$m$$.