Definite integral involving exponential and logarith function

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Asked 25 days ago

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Working with Dilogarimth function, we get the following definite integral

$$\int_0^{\infty}\frac{t^2\,\ln^{n}(t)}{(1-e^{x\,t})(1-e^{y\,t})}\,dt$$ with $n=1,2,3,...$ and $x,y>0$.

I wonder if is possible write in terms of elementary functions or (more probably) in terms of special function.

Any help is welcomed.

Edit: I have added $t^2$ at numerator, in order to convergence.

edited Jun 14 at 9:30

asked Jun 13 at 18:09

popi

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$\begingroup$ The integral $\int_0^{\infty}\frac{t^{\alpha}}{(1-e^{x\,t})(1-e^{y\,t})}\,dt$ is equal to $$\sum_{n=0}^\infty\sum_{k=0}^\infty\frac{\Gamma(1+\alpha)}{\left((n+1)x+(k+1)y\right)^{1+\alpha}} $$ If this has a closed form then differentiate it $n$ times and set $\alpha=0$ to get the desired result. $\endgroup$
– Zima
Commented Jun 13 at 18:26
3

$\begingroup$ The integral does not seem to converge since it blows up near $t=0$ $\endgroup$
– mattTheMathLearner
Commented Jun 14 at 2:55
$\begingroup$ Yes, it's true. I forgot a "t" at the numerator. $\endgroup$
– popi
Commented Jun 14 at 8:12
1

$\begingroup$ Still does not converge. Try looking at the series expansion near $t=0$. $\endgroup$
– mattTheMathLearner
Commented Jun 14 at 8:48
1

$\begingroup$ Assuming $n=1,x=1,y=2$, following the approach suggested by @Zima, differentiating the summation once and setting $\alpha = 2$, I get the numerically evaluated sum to be $-0.340481$ which matches with the numerically evaluated integral. $\endgroup$
– mattTheMathLearner
Commented Jun 14 at 16:36

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