Question

Show that
\[\int_0^\infty \frac{dx}{1+x^n}=\frac{\pi}{n}\csc\frac{\pi}{n}\]
for \(n\in\mathbb{N}\backslash\{1\}\).

Edit: another relationship that appears to be true:
\[\int_0^\infty \frac{dx}{1+x^n}=\frac{1}{n}\Gamma\left(\frac{1}{n}\right)\Gamma\left(1-\frac{1}{n}\right)=\frac{1}{n}\mathrm{B}\left(\frac{1}{n},1-\frac{1}{n}\right)\]

Answer 1

The identities
\[\Gamma(x)\Gamma(-x)=-\frac{\pi}{x\sin(\pi x)}\quad\text{and}\quad\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin (\pi x)}\]
seem to shed some light on the relation above.
(Identities 41 and 42 shown here: http://mathworld.wolfram.com/GammaFunction.html )

Answer 2

Applying the residue theorem and using a half-circle contour over the upper-half plane seems an okay strategy, but calculating all those residues... Not sure if induction would help, but then again, I have yet to try.

Answer 3

\[\int\limits_{0}^{\infty}\frac{dx}{1+x^2}=\arctan(x)|_0^{\infty}=\frac{\pi}{2}=\frac{\pi}{2} \csc(\frac{\pi}{2})\]
So it works for n=2
now assume it works for some integer k
that is we have \[\int\limits_{0}^{\infty}\frac{dx}{1+x^k}=\frac{\pi}{k} \csc(\frac{\pi}{k})\]
now let's show this is true for k+1
that is we want to show
\[\int\limits_{0}^{\infty}\frac{dx}{1+x^{k+1}}=\frac{\pi}{k+1} \csc(\frac{\pi}{k+1})\]
hmm....
I wonder if is possible to prove it by induction
actually stuck right now here

Answer 4

great job

Answer 5

Here's something we might be able to use...

For \(|x|<1\),
\[\frac{1}{1+x^n}=\frac{1}{1-(-x^n)}=\sum_{k=0}^\infty(-1)^kx^{nk}\]
For \(|x|>1\),
\[\frac{1}{1+x^n}=\frac{1}{x^n}\frac{1}{1-(-x^{-n})}=\frac{1}{x^n}\sum_{k=0}^\infty (-1)^kx^{-nk}=\sum_{k=0}^\infty (-1)^k x^{-n(k+1)}\]
which we know converges because the ratio test gives us this result:
\[\lim_{k\to\infty}\left|\frac{(-1)^{k+1} x^{-n(k+2)}}{(-1)^k x^{-n(k+1)}}\right|=|x|^{-n}<1~~\iff~~1<|x|^n~~\implies~~1<|x|\]
So,
\[\begin{align*}
\int_0^\infty \frac{dx}{1+x^n}

&=\left\{\int_0^1+\int_1^\infty\right\}\frac{dx}{1+x^n}\\\\

&=\sum_{k=0}^\infty \left\{\int_0^1 (-1)^kx^{nk}\,dx+\int_1^\infty (-1)^kx^{-n(k+1)}\,dx\right\}\\\\

&=\sum_{k=0}^\infty \left\{(-1)^k\left[\frac{x^{nk+1}}{nk+1}\right]_{0}^{1}+(-1)^k\left[\frac{x^{-n(k+1)+1}}{-n(k+1)+1}\right]_1^\infty\right\}\\\\

&=\sum_{k=0}^\infty \left\{\frac{(-1)^k}{nk+1}-\frac{(-1)^{k}}{1-n(k+1)}\right\}\\\\

&=\left(1-\frac{1}{1-n}\right)+\left(-\frac{1}{n+1}+\frac{1}{1-2n}\right)\\
&\quad\quad+\left(\frac{1}{2n+1}-\frac{1}{1-3n}\right)+\left(-\frac{1}{3n+1}+\frac{1}{1-4n}\right)+\cdots\\\\

&=1-\frac{2}{1-n^2}+\frac{2}{1-4n^2}-\frac{2}{1-9n^2}+\cdots\\\\

&=\sum_{k=0}^\infty \frac{(-2)^k}{1-(kn)^2}
\end{align*}\]
Hmm, this kinda resembles the Basel problem.

Answer 6

Actually, there might be a flaw in this work somewhere along the line, since
\[\lim_{k\to\infty}\left|\frac{(-2)^{k+1}}{1-(k+1)^2n^2}\times\frac{1-k^2n^2}{(-2)^k}\right|=2\lim_{k\to\infty}\left|\frac{\frac{1}{k^2}-n^2}{1-\left(\frac{k+1}{k}\right)^2n^2}\right|=2>1\]
and so the series appears to diverge by the ratio test...

Answer 7

Apparently, the contour method seems to work out nicely when you use a contour that surrounds only one of the complex roots, as shown here: http://math.stackexchange.com/questions/247866/show-that-int-0-infty-frac11xn-dx-frac-pi-n-sin-pi-n-wh

But I think I'm more interested in a real method of approaching this.

Answer 8

sub \(u = \dfrac{1}{1+x^n} \implies x = \left(\dfrac{1}{u}-1\right)^{1/n}\)

the integral becomes :
\[\begin{align}\int_0^\infty \frac{dx}{1+x^n} &= \frac{1}{n} \int_0^1 u^{-\frac{1}{n}}(1-u)^{\frac{1}{n}-1} \,du\\~\\
&=\frac{1}{n} B\left(1-\frac{1}{n}, ~\frac{1}{n}\right)\\~\\
&= \frac{1}{n}\dfrac{\Gamma(1-\frac{1}{n}) \Gamma(\frac{1}{n})}{\Gamma(1)}\\~\\
&=\frac{1}{n} \frac{\pi}{\sin(\frac{\pi}{n})}

\end{align}\]
for a more general case see
http://openstudy.com/users/ganeshie8#/updates/54a7145ce4b054f0c3b7e558

Answer 9

induction idea seems new xD

Answer 10

Oh yeah I remember seeing that question, totally slipped my mind