# Evaluating $\int_0^\infty \sqrt{\frac{x}{e^x-1}}dx$ in terms of special functions

Introduction: I've been studying integrals of the form $$\int_0^\infty \frac{x^a}{(e^x-1)^b}dx$$ where a and b are real parameters. I've been able to find closed forms for the integral in terms of the Riemann Zeta function, the Gamma function and the Polygamma functions provided the integral converges when at least one of the parameters is a positive integer.

I then thought of generalizing this to the case where both a and b are non-integers. As a start I considered the integral $$\int_0^\infty \sqrt\frac{x}{e^x-1}dx$$ Using substitution methods and integration by parts I deduced that $$\int_0^\infty \sqrt\frac{x}{e^x-1}dx=\int_1^\infty \frac{\sqrt{\ln x}}{x\sqrt{x-1}}dx=\int_0^1 \sqrt\frac{-\ln x}{x(1-x)} dx=2\sqrt2\int_0^1 \sqrt\frac{-\ln x}{1-x^2}dx$$ $$=\sqrt2\int_0^1 \frac{\arcsin x}{x\sqrt{-\ln x}}dx=\sqrt2\int_0^\infty \frac{\arcsin(e^{-x})}{\sqrt x} dx=2\sqrt2\int_0^\infty \arcsin(e^{-x^2})dx$$ $$=2\sqrt2\int_0^\frac{\pi}{2} \sqrt{-\ln(\sin x)} dx$$

I tried different methods on each integral but pretty much nothing worked. With each method not shown here the integral got much more complicated. The last integral looks similar to the Clausen's integral but I wasn't able to establish a relation between them. It is of course possible to represent the value of the integral as an infinite series but my question is that does anybody know a method to evaluate this integral in terms of well-known special functions or mathematical constants? Thanks in advance.

## 2 Answers

A partial answer.

I'm sure you already know the Lerch transcendent, a special function which may initially be defined as $$\Phi(z,s,a):=\sum_{k=0}^\infty\frac{z^k}{(a+k)^s}, \quad a>0,\Re s>1,|z|<1.$$ It admits the following integral representation, which you obtain by expanding the integrand as a powers series of $z$: $$\Phi(z,s,a)=\int_0^{\infty}\frac{x^{s-1}e^{-ax}}{1-ze^{-x}}{\rm d}x.$$ By differentiation with respect to $z$, you get $$\partial_z^r\Phi(z,s,a)=(-1)^r\int_0^{\infty}\frac{x^{s-1}e^{-(a+r)x}}{(1-ze^{-x})^{r+1}}{\rm d}x.\tag1$$ This shows the level of complexity of your family of integrals: fractional calculus.

For example, from $(1)$, you have

$$\int_0^\infty \sqrt\frac{x}{e^x-1}dx= \color{blue}{\left(-\partial_z\right)^{\large \! -1/2}\Phi\left(1,3/2,1\right). }$$

I would be very surprised to learn that we have more than the above expression.

• Well apparently things got complicated quite quickly. So is differentiation under the integral sign also valid for fractional derivatives? That's new to me. Out of curiosity, is it often used to evaluate integrals with fractional powers? Anyway, this is the kind of answer I was looking for so thank you. – user203664 Dec 29 '14 at 1:43

Starting from the general $$~\displaystyle\int_0^\infty\frac{x^a}{e^x-u}dx~=~\frac{\Gamma(a+1)\cdot\text{Li}_{a+1}(u)}u,~$$ we can then

deduce, by way of repeated differentiation under the integral sign with regard to u,

that $$~\displaystyle\int_0^\infty\frac{x^a}{(e^x-u)^b}dx~=~(-1)^b\cdot\Gamma(a+1)\cdot{\large\partial}^b_u~\bigg[\dfrac{\text{Li}_{a+1}(u)}u\bigg],~$$ for all real $$a>-1$$

and $$u\in\mathbb C$$, with $$b\in\mathbb N.~$$ However, in this case, $$~b=\dfrac12~$$ is fractional, so, if you are

looking for a closed form, you will have to appeal to the concept of fractional calculus,

which has already been mentioned above, though in relation to a lesser-known special

function.