I have found a proof using complex analysis techniques (contour integral, residue theorem, etc.) that shows $$\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$$ for $n\in \mathbb{N}^+\setminus\{1\}$

I wonder if it is possible by using only real analysis to demonstrate this "innocent" result?

Edit A more general result showing that $$\int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx} = \frac{\pi}{b \sin(\pi{a}/b)}, \qquad 0 < a <b$$ can be found in another math.SE post


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

$$ =\int_{0}^{\infty}\int_{0}^{\infty}e^{-t}e^{-tx^{n}}\ dx\ dt =\frac{1}{n}\int_{0}^{\infty}\int_{0}^{\infty}e^{-t}e^{-u}\Big(\frac{u}{t}\Big)^{\frac{1}{n}-1}\frac{1}{t}\ du\ dt $$

$$ =\frac{1}{n}\int_{0}^{\infty}t^{-\frac{1}{n}}e^{-t}\int_{0}^{\infty}u^{\frac{1}{n}-1}e^{-u}\ du\ dt =\frac{1}{n}\int_{0}^{\infty}t^{-\frac{1}{n}}e^{-t}\ \Gamma\Big(\frac{1}{n}\Big)\ dt $$

$$ =\frac{1}{n}\ \Gamma\Big( 1-\frac{1}{n}\Big)\Gamma\Big(\frac{1}{n}\Big) =\frac{\pi}{n}\csc\Big(\frac{\pi}{n}\Big) $$

  • 3
    $\begingroup$ :nice approach but why ? $$\frac{1}{n}\ \Gamma\Big( 1-\frac{1}{n}\Big)\Gamma\Big(\frac{1}{n}\Big)=\frac{\pi}{n}\csc\Big(\frac{\pi}{n}\Big)$$ $\endgroup$ – M.H Jun 29 '13 at 15:00
  • 1
    $\begingroup$ @MaisamHedyelloo $\Gamma(z)\Gamma(1-z)=\dfrac{\pi}{\sin (\pi z)}$, for $0<z<1$. $\endgroup$ – Cortizol Jun 29 '13 at 15:02
  • $\begingroup$ @Cortizol :thanks $\endgroup$ – M.H Jun 29 '13 at 15:04
  • 3
    $\begingroup$ @RandomVariable This is by $\Gamma(z)\Gamma(1-z)=\pi/\sin(\pi z)$, but how do you get it using only real analysis? $\endgroup$ – Sungjin Kim May 31 '14 at 14:14
  • 3
    $\begingroup$ @i707107 By using the Weierstrass infinite product representation of the gamma function, which can be derived without using complex analysis from Euler's limit definition of the gamma function. planetmath.org/eulerreflectionformula $\endgroup$ – Random Variable May 31 '14 at 14:30

Here is another method. Let $\varphi: x\mapsto \displaystyle\int_0^{+\infty}\dfrac{dt}{1+t^x}$ and $\psi:x\mapsto\dfrac{\dfrac {\pi}{x}}{\sin\left(\dfrac{\pi}{x}\right)}$

You can easily prove that $\varphi$ and $\psi$ are both defined and continuous on $\mathscr I=\mathbb ]1,+\infty[$.

Let $\mathscr{A}=\left\{\dfrac pq : p\in2\mathbb Z, q\in2\mathbb Z+1 \right\}$ and $\mathscr A^*_+=\mathscr A \cap \mathscr I$. Then $\mathscr A^*_+$ is a dense subset of $\mathscr I$.

Now let $p$ be an even integer and $q$ an odd one such that $p>q>0$. We have :

$$\varphi\left(\frac{p}{q}\right)=\int_0^{+\infty}\frac{dt}{1+t^\frac{p}{q}}=\int_0^{+\infty}\frac{qu^{q-1}}{1+u^p}du= \frac{q}{2}\int_{-\infty}^{+\infty}\frac{u^{q-1}}{1+u^p}du=\frac{q}{2}\lim_{t\rightarrow+\infty}\int_{-t}^t\frac{u^{q-1}}{1+u^p}du$$

We can write : $\displaystyle{\frac{u^{q-1}}{1+u^p}=\sum_{k=0}^{p-1}\frac{a_k}{u-b_k}}$ with $\displaystyle{b_k=e^{i\frac{(2k+1)\pi}{p}}}$ and $\displaystyle{a_k=\frac{-b_k^q}{p}=-\frac{e^{i\frac{(2k+1)\pi q}{p}}}{p}}$

Now let $x$ be a real number such that $\sin(x)\neq 0$.

We can then write : $\displaystyle{\frac{1}{u-e^{ix}}=\frac{u-\cos(x)+i\sin(x)}{u²-2u\cos(x)+1}}$

Now if $t>0$ we get : $\displaystyle{\int_{-t}^t\frac{u-\cos(x)}{u²-2u\cos(x)+1}du=\frac{1}{2}\ln\left(\frac{t²-2t\cos(x)+1}{t²+2t\cos(x)+1}\right)}$ and this integral tends to $0$ as $t$ tends to $+\infty$.

We have as well : $\displaystyle{\int_{-t}^t\frac{\sin(x)}{u²-2u\cos(x)+1}du=\arctan\left(\frac{t-\cos(x)}{\sin(x)}\right)+\arctan\left(\frac{t+\cos(x)}{\sin(x)}\right)}$ and this integral tends to $\pi$ if $\sin(x)>0$ and $-\pi$ if $\sin(x)<0$ (when $t$ tends to $+\infty$).

So we get : $$\lim_{t\to +\infty} \int_{-t}^t \dfrac{du}{u-e^{ix}}=\left\{\begin{array}{lr} i\pi & \text{if}\ \sin(x)>0\\ -i\pi & \text{if}\ \sin(x)<0\end{array}\right.$$

Now let's go back to our little integral :

$$\dfrac q2 \lim_{t\to +\infty}\int_{-t}^t \dfrac{u^{q-1}}{1+u^p} du=i\pi\dfrac q2\left(\sum_{k=0}^{\frac p2-1}a_k-\sum_{k=\frac p2}^{p-1} a_k\right)=-q\pi\mathrm{Im}\left(\sum_{k=0}^{\frac p2-1} a_k\right) \ (\text{because}\ a_k=\overline{a_{p-1-k}})$$

But this last sum is just a simple geometric sum :

$$\sum_{k=0}^{\frac p2-1} a_k=-\dfrac 1p e^{i\pi\frac qp}\frac{1-e^{i\pi q}}{1-e^{2i\pi\frac qp}}=-\frac{i}{p\sin\left(\pi\frac qp\right)}$$

And finally we get :

$$\varphi\left(\frac pq\right)=-q\pi\left(-\frac1{p\sin\left(\pi\frac qp\right)}\right)=\frac{\dfrac{\pi}{\frac pq}}{\sin\left(\dfrac{\pi}{\frac pq}\right)}=\psi\left(\frac pq\right)$$

$\varphi$ and $\psi$ are both continuous and they agree on the dense subset $\mathscr A^*_+$ of $\mathscr I$. Hence they're equal.

  • $\begingroup$ It is a bit longer but I like this answer because it is self contained; that is, it does not rely on the reflection property of $\Gamma(t)$. $\endgroup$ – Mark Fischler Mar 31 '15 at 19:37

We can use the geometric series $\frac{1}{1-x}=\sum_{k=0}^\infty x^n$ for $|x|<1$ to evaluate: \begin{eqnarray} \int_0^\infty\frac{1}{1+x^n}dx&=&\int_0^1\frac{1+x^{n-2}}{1+x^n}dx\\ &=&\sum_{k=0}^\infty(-1)^k\int_0^1(1+x^{n-2})x^{nk}dx\\ &=&\sum_{k=0}^\infty(-1)^k\left(\frac{1}{nk+1}+\frac{1}{nk+n-1}\right)\\ &=&\sum_{k=-\infty}^\infty(-1)^k\frac{1}{nk+1}\\ &=&\frac{\pi}{n\sin\frac{\pi}{n}}. \end{eqnarray}

  • $\begingroup$ The upper limit is infinity, not $1$. $\endgroup$ – Tunk-Fey Jul 23 '14 at 19:07
  • 1
    $\begingroup$ Monotone Convergence Theorem at line 2? $\endgroup$ – Rrjrjtlokrthjji Oct 14 '14 at 19:53
  • $\begingroup$ In line 1: $\int_0^\infty \frac {1} {1+x^2} = \int_0^1 \frac {1} {1+x^2}+\int_1^\infty \frac {1} {1+x^2}$ and in the second integral $x\to\frac {1}{y}$ $\endgroup$ – Dr. Wolfgang Hintze Apr 2 '19 at 9:48

In general, let $$y=\dfrac{1}{1+x^b}\quad\Rightarrow\quad x=\left(\dfrac{1-y}{y}\right)^{\frac1b}\quad\Rightarrow\quad dx=-\left(\dfrac{1-y}{y}\right)^{\frac1b-1}\ \dfrac{dy}{by^2}\ ,$$ then \begin{align} \int_0^\infty\dfrac{x^{a-1}}{1+x^b}\ dx&=\int_0^1 y\left(\dfrac{1-y}{y}\right)^{\large\frac{a-1}b}\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\\&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy, \end{align} where the integral in RHS is Beta function. $$ \text{B}(x,y)=\int_0^1t^{\ x-1}\ (1-t)^{\ y-1}\ dt=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}. $$ Hence \begin{align} \int_0^\infty\dfrac{x^{a-1}}{1+x^b}\ dx&=\frac1b\int_0^1y^{1-\frac{a}{b}-1}(1-y)^{\frac{a}{b}-1}\ dy\\&=\frac1b\cdot\Gamma\left(1-\frac{a}{b}\right)\cdot\Gamma\left(\frac{a}{b}\right)\\&={\color{blue}{\frac{\pi}{b\sin\left(\frac{a\pi}{b}\right)}}}. \end{align} The last part uses Euler's reflection formula for Gamma function provided $0<a<b$. Thus $$ \int_0^\infty\dfrac{1}{1+x^n}\ dx=\color{blue}{\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}}. $$


  • 1
    $\begingroup$ What does Q.E.D. mean? $\endgroup$ – jeanne clement Jan 12 '17 at 11:20
  • $\begingroup$ QED is Latin for quod erat demonstrandum meaning "that which was to be demonstrated, shown, or proved." Less formally, one often hears QED as corresponding to "Quite Easily Done." $\endgroup$ – omegadot Nov 7 '17 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.