Hot Integral-3

Really, I am a fan of you and your problems(especially this Hot Integral series).

This is our problem -

$\displaystyle \int \limits_{0}^{\infty}{\frac{4{t}^{2}{e}^{-3t}}{2-{e}^{-t}} dt}$

$\displaystyle = 2\int \limits_{0}^{\infty}{\frac{{t}^{2}{e}^{-3t}}{1-{e}^{-t}/2} dt}$

Now, we will use this famous series - $\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty}{{x}^{n}}$

And we get,

$\displaystyle \sum_{n=0}^{\infty}{\frac{1}{{2}^{n}} \int \limits_{0}^{\infty}{{t}^{2}{e}^{-3t} {e}^{-nt} dt}}$

Now, you may call that integral a Laplace transform of ${t}^{2}$ or even Gamma Function and get its value as -

$\displaystyle \frac{2!}{{(n+3)}^{3}}$

And substituting that in our sum, we get -

$\displaystyle \frac{1}{2} \sum_{n=0}^{\infty}{\frac{1}{{2}^{n}{(n+3)}^{3}}}$

Well, what's that? Polylogarithm equivalent for Hurwitz zeta function? Yeah, and we have a special name for that(as far as my knowledge says) - Lerch Transcendent . And all the other "nearly same functions" become special cases of this general function. Our answer is equal to $\displaystyle \Phi \left(\frac{1}{2}, 3, 3 \right)$ .

But let's do something different.

Again consider our "basic" series -

$\displaystyle 1 + x + {x}^{2} + \cdots = \frac{1}{1-x}$

Multiplying by ${x}^{2}$ and then integrate both sides,

$\displaystyle \sum_{k=3}^{\infty}{\frac{{x}^{k}}{k}} = \int{\frac{{x}^{2}}{1-x}}$

And similarly doing this, we get,

$\displaystyle \sum_{k=0}^{\infty}{\frac{{x}^{k}}{{(k+3)}^{3}}} = \frac{1}{{x}^{3}} \int{\frac{1}{x} \int{\frac{1}{x} \int{\frac{{x}^{2}}{1-x}}}}$

Now, it's easy to do this integration and we get

$7\zeta(3) - \frac{17}{4} + \frac{4}{3}{log}^{3}(2) - \frac{2}{3}{\pi}^{2} log(2)$

This is Lerch Transcendent. Its integral representation :

$\displaystyle \Phi(z,s,a) = \frac{1}{\Gamma s} \int\limits_0^{\infty} \frac{t^{s-1}e^{-at}}{1 - ze^{-t}} dt$

Comparing we get $z = 1/2 , s = 3 , a = 3$

Solve for $\Phi(\frac{1}{2},3,3)$

= $\displaystyle 7\zeta(3) - \frac{17}{4} + \frac{4}{3}\log^3 2 - \frac{2}{3}\pi^2\log^2 2$

$\displaystyle \frac{1}{2}\int_{0}^{\infty}\frac{t^{2}e^{-3t}}{1-\frac{e^{-t}}{2}}dt$

$\displaystyle = \frac{1}{2}\sum_{r=0}^{\infty}\int_{0}^{\infty}\frac{t^{2}e^{-(r+3)t}}{2^{r}}dt$

Here we used the fact that $\displaystyle \frac{1}{1-x} = \sum_{r=0}^{\infty}x^{r}$ , when $-1<x<1$

Substituting $(r+3)x=t$ and using Gamma Function we have:-

$\displaystyle 2\sum_{r=0}^{\infty}\frac{1}{(r+3)^{3}2^{r+1}} = 8\sum_{r=0}^{\infty}\frac{1}{(r+3)^{r+3}2^{r+3}} = 8\left(\text{Li}_{3}\left(\frac{1}{2}\right)-\frac{17}{32}\right)$

Now substituting the value of $\text{Li}_{3}(\frac{1}{2})$ we have our answer as :-

$\displaystyle 7\zeta(3) -\frac{17}{4} + \frac{4}{3}\ln^{3}(2)-\frac{2\pi^{2}}{3}\ln(2)$

Notation:- Here $\text{Li}_{3}(\cdot)$ denotes the Trilogarithm function .