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I have to show that $$\int_{0}^{\infty}\frac{x^4e^x}{(e^x-1)^2}dx \approx \int_{0}^{a} \frac{x^4e^x}{(e^x-1)^2}dx$$ for large values of $a$. This can be done by showing that $$\int_{a}^{\infty}\frac{x^4e^x}{(e^x-1)^2}dx$$ is exponentially small.

However, I am having trouble finding a way to integrate $$\int_{a}^{\infty}\frac{x^4e^x}{(e^x-1)^2}dx$$ analytically.

I was given a suggestion to think about making an approximation to $x^4$ when integrating over a region where $x$ is large, but I'm completely lost.

Any suggestions is greatly appreciated.

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  • $\begingroup$ What about using $x^{4} \le e^{x}$ for $x \ge 0$? Then you can bound your integral from above. It seems to come out pretty nicely. $\endgroup$ – mattos Mar 21 '18 at 4:50
  • $\begingroup$ What values of $a$ do you consider large? Do you have access to software like Mathematica which can calculate the integral to any arbitrary precision? $\endgroup$ – JimB Mar 21 '18 at 5:09
  • $\begingroup$ Hi Mattos, thanks for the suggestion, but I'm not exactly sure what you mean by being able to bound the integral from above using the given inequality. $\endgroup$ – Plana Mar 21 '18 at 5:11
  • $\begingroup$ JimB, I unfortunately do not have access to such software. The question just states "for large values of $a$," so I wouldn't know how to answer your first question. $\endgroup$ – Plana Mar 21 '18 at 5:14
  • $\begingroup$ I came up with perhaps a foolish attempt at the solution inspired from Mattos' suggestion which would be to say that $1< x^4 << e^x$ for large values of $x$, so the integral would simplify to $\int_{a}^{\infty} \frac{e^x}{(e^x-1)^2}dx$ Which comes out to be $\frac{1}{e^a -1} \approx e^{-a} $ but I can't stand for the legitimacy of this solution... $\endgroup$ – Plana Mar 21 '18 at 5:21
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We have that for $a>\ln(2)$, $$\int_{a}^{\infty}\frac{x^4e^x}{(e^x-1)^2}dx=\int_{a}^{\infty}\frac{x^4e^{-x/2}}{e^{x/2}(1-e^{-x})^2}dx< \frac{\int_{0}^{\infty} x^4 e^{-x/2}dx}{e^{a/2}(1-e^{-\ln(2)})^2}=\frac{2^2\cdot 2^5\cdot 4!}{e^{a/2}}$$ which implies that the LHS goes to zero exponentially as $a$ goes to $+\infty$.

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For sufficiently large $x$, $$\frac{x^4e^x}{(e^x-1)^2}=\frac{x^4}{(e^\frac{x}{2}-e^\frac{-x}{2})^2}$$ $$=\frac{x^4}{(\sum_{n=1}^{\infty} \frac{x^{2n}}{2^{2n}(2n)!})^2} <\frac{x^4}{\sum_{n=1}^{\infty} \frac{x^{4n}}{(2^{2n}(2n)!)^2}}$$ $$<\frac{x^4}{\frac{x^8}{k}}$$ $$=\frac{k}{x^4}$$ where $k$ is some constant. Thus, $$0<\frac{x^4e^x}{(e^x-1)^2}<\frac{k}{x^4}$$ Integrating both sides gives the required answer.

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