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.
Best Answer
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.