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I was trying to compute the value of the definite integral

$$\ I = \int_{0}^{\infty} \frac{x^ne^x}{e^{2x}+1}dx$$

I noticed that the Gamma function would be particularly relevant here, since after all

$$\Gamma(n+1) = \int_{0}^{\infty} x^n e^{-x}dx$$

To obtain something that resembles the Gamma function, I divided the fraction by $\ e^{2x}$

$$\ I = \int_{0}^{\infty} \frac{x^ne^{-x}}{e^{-2x}+1}dx$$

And then noticed that

$$\frac{1}{1+e^{-2x}} = 1 - e^{-2x} + e^{-4x} - e^{-6x} + e^{-8x} ..... $$

Evaluating each term, we get:

$$\ I = \int_0^{\infty} x^ne^{-x}dx - \int_0^{\infty} x^ne^{-3x}dx + \int_0^{\infty} x^ne^{-5x}dx - \int_0^{\infty} x^ne^{-7x}dx.....$$

$$\ I = n! - \frac{n!}{3^n} + \frac{n!}{5^n} - \frac{n!}{7^n}..... = n!\cdot (1-\frac{1}{3^n}+\frac{1}{5^n} -\frac{1}{7^n}.....)$$

This immediately reminded me of the Riemann Zeta function, but I am unable to express the final result in terms of it. I know that

$$\ 1 +\frac{1}{3^n}+\frac{1}{5^n}+\frac{1}{7^n}+\dots = \zeta(n) - \frac{1}{2^n}\zeta(n)$$

But I was unable to find the alternating difference, which is the expression in the final result. Can someone please help me with that? Thank you for reading.

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    I think you just derived https://en.m.wikipedia.org/wiki/Dirichlet_beta_function. – Bob Dobbs Dec 19 '23 at 18:21
  • @BobDobbs Whoa, thanks! That's cool, didn't know about this. These special functions sure are handy things. –  Dec 19 '23 at 18:32

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One of my older comments should help here (as in Bob Dobbs' comment) : \begin{align} I(m):=\int_0^\infty \frac{t^m}{2\,\cosh(t)}dt&=\int_0^\infty \frac{t^m\;e^{-t}}{1+e^{-2t}}dt\\ &=\int_0^\infty \sum_{k=0}^\infty (-1)^k\;t^m\;e^{-(2k+1)t}\;dt\\ &=\sum_{k=0}^\infty\frac {(-1)^k}{(2k+1)^m}\int_0^\infty u^m\;e^{-u}\frac{du}{2k+1}\;\\ &=\Gamma(m+1)\sum_{k=0}^\infty\frac {(-1)^k}{(2k+1)^{m+1}}\\ &=\Gamma(m+1)\,\beta(m+1)\\ \end{align}

where $\beta$ is the Dirichlet $\beta$ function.

Raymond Manzoni
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  • Whoa, thanks a lot! Your answer looks very unexpected, wouldn't expect such a complicated integral to evaluate into such a nice looking integer! –  Dec 19 '23 at 18:36
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    Let's be honest you found it yourself :-) and just needed a correct term for the alternating series! Excellent continuation anyway, (for the reader I used the substitution $,u:=(2k+1),t$) – Raymond Manzoni Dec 19 '23 at 18:39
  • Believe me, I didn't. The truth is, I was actually studying the integral $\int_0^{\frac{\pi}{4}} cos(ln(tanx))dx$. When you substitute $\ tanx = t$ and then $\ lnt = u$, followed by some rearranging, we get the integral of $\int_0^{\infty} \frac{e^xcos(x)}{e^{2x}+1}dx$. Then I was merely playing around, putting the Taylor series of cos(x) about 0 into the integral, and that led me to thinking about integrals of the form $\frac{x^ne^x}{e^{2x}+1}$. –  Dec 19 '23 at 19:01
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    In my comments I indicate too $\displaystyle \frac 1{\cosh(t)}=\int_0^\infty \frac {\cos(t,x)}{\cosh(\pi,x/2)};dx$ that should help you to simplify your $\displaystyle\int_0^{\infty} \frac{e^xcos(x)}{e^{2x}+1}dx,$ formula! You may too try directly Alpha – Raymond Manzoni Dec 19 '23 at 19:30
  • Do you mind sharing your thoughts on this follow-up question? –  Dec 19 '23 at 21:29
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    For a direct derivation you could use the relation $(7)$ from Mathworld between $\beta(2n+1)$ and the Euler number $E_{2n}$ : $$(-1)^n\beta(2n+1)=\frac{ E_{2n}}{2(2n)!}\left(\frac {\pi}2\right)^{2n+1}$$ and conclude with the $E_n$ exponential generating function at $t=\frac{\pi}2$ (the odd terms being $0$) : $$\frac 1{\cosh t}=\sum_{n=0}^\infty \frac {E_n}{n!}t^n$$ Answering yourself is fine! Cheers, – Raymond Manzoni Dec 19 '23 at 22:14
  • As indicated by X-Rui $\beta(n)$'s limit is $1$ so that this is rather a formal derivation using generating functions (i.e. not requiring convergence). A better idea could be to study the limit as $;\displaystyle t\to \frac {\pi}2, t<\frac {\pi}2$ – Raymond Manzoni Dec 19 '23 at 23:30