Prove $\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$ using real analysis techniques only

Question

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

Answer 1

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

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

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

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

Answer 2

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 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.

Answer 3

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 $$ \int_0^\infty\dfrac{1}{1+x^n}\ dx=\color{blue}{\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}}. $$

QED

Answer 4

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}

Answer 5

Euler originally derived this by partial fraction decomposition. The derivation can be found on this site: http://www.17centurymaths.com/contents/integralcalculus.html. For the later steps, see Vol 1, Sec 1, Chap 8. For the partial fraction decomposition, see Vol 1, Sec 1, Chap 1. $$ \int_0^{\infty} \frac{x^{m-1}}{1+x^n} dx = \frac{\pi}{n \sin (m \pi / n ) } $$

1. The roots of $x^n + 1 = 0 $ is $ x = e^{i \varphi_k} $ where

$$ \begin{eqnarray} \varphi_k &=& \pm \frac{2k-1}{n} \pi ~~\left(n:\textrm{even}, k = 1,2, .., \frac{n}{2}\right)\\ \varphi_k &=& \pm \frac{2k-1}{n} \pi, \pi ~~\left(n:\textrm{odd}, k = 1,2, ..,\frac{n-1}{2} \right) \end{eqnarray} $$

2. Combining conjugates, $x^n+1$ can be factored as $$ \begin{eqnarray} x^n + 1 &=& \prod_k^{\frac{n}{2}}(x^2-2 x \cos \varphi_k + 1) ~~(n:\textrm{even}) \\ x^n + 1 &=& (x+1) \prod_k^{\frac{n-1}{2}}(x^2-2 x \cos \varphi_k + 1) ~~(n:\textrm{odd}) \end{eqnarray} $$

3. Suppose $x^n + 1 = (x^2 - 2 x \cos \varphi + 1) S(x)$. Consider the following partial fractions. $$ \frac{x^{m-1}}{x^n+1} = \frac{A x + B}{x^2-2x \cos \varphi + 1} + \frac{V(x)}{S(x)} $$

where A,B are real numbers

4. multiply $x^n + 1$ to both sides and substitute $x = e^{i \varphi}$.

$$ e^{i (m-1) \varphi} = (A e^{i \varphi} + B)S(e^{i \varphi}) $$

This looks like an equation with 2 unknown variables, but it is actually solvable because it is an equation of complex numbers and we constrain A and B to be real numbers.

5. define following notation for convenience. $$ \begin{eqnarray} a \cdot b &=& \frac{\overline{a} b + a \overline{b}}{2} \\ a \times b &=& \frac{\overline{a} b - a \overline{b}}{2 i} \end{eqnarray} $$

where $\overline{a}$ is a complex conjugate of $a$. Following relations can be found

$$ \begin{eqnarray} a \cdot a &=& |a|^2 \\ a \times a &=& 0 \\ a \cdot a e^{i \theta} &=& |a|^2 \cos \theta \\ a \times a e^{i \theta} &=& |a|^2 \sin \theta \\ e^{i \theta_1} \cdot e^{i \theta_2} &=& \cos (\theta_2 - \theta_1) \\ e^{i \theta_1} \times e^{i \theta_2} &=& \sin (\theta_2 - \theta_1) \end{eqnarray} $$

6. let $s = S(e^{i \phi})$, Using relations in 5, 4 can be solved as

$$ \begin{eqnarray} A &=& \frac{s \times e^{i (m-1) \varphi}}{|s|^2 \sin \varphi} \\ B &=& \frac{1}{|s|^2 \sin \varphi} \left(s\cdot e^{i (m-1) \varphi} \sin \varphi - s \times e^{i (m-1) \varphi} \cos \varphi \right) \end{eqnarray} $$

7. find s as follows. $$ \begin{eqnarray} s &=& \frac{x^n + 1}{x^2 - 2 x \cos \varphi + 1} \Big|_{x \to x^{i \varphi}} \\ &=& \frac{n}{2} \frac{e^{i (n-1) \varphi}}{i \sin \varphi} \end{eqnarray} $$

$$ \begin{eqnarray} s \cdot e^{i (m-1) \varphi} &=& \frac{n \sin (n-m) \varphi}{2 \sin \varphi} \\ s \times e^{i (m-1) \varphi} &=& \frac{n \cos (n-m) \varphi}{2 \sin \varphi} \end{eqnarray} $$

9. Using 8, from 6,

$$ \begin{eqnarray} A &=& \frac{2 \cos (n-m) \varphi}{n} \\ B &=& - \frac{2 \cos (n-m + 1) \varphi}{n} \end{eqnarray} $$ Using $n \varphi = (2k-1) \pi$, \begin{eqnarray} A &=& - \frac{2 \cos m \varphi}{n} \\ B &=& \frac{2 \cos (m - 1) \varphi}{n} \end{eqnarray}

10. Thus partial fraction decomposition of $\frac{x^{m-1}}{x^n+1}$ is

$$ \begin{eqnarray} \frac{x^{m-1}}{x^n+1} &=& \frac{2}{n} \sum_{k=1}^{\frac{n}{2}} \frac{- x \cos m \varphi_k + \cos (m-1) \varphi_k }{x^2 - 2 x \cos \varphi_k + 1} ~~ (n:\textrm{even}) \\ \frac{x^{m-1}}{x^n+1} &=& \frac{2}{n} \sum_{k=1}^{\frac{n-1}{2}} \frac{- x \cos m \varphi_k + \cos (m-1) \varphi_k }{x^2 - 2 x \cos \varphi_k + 1} + \frac{1}{n} \frac{(-1)^{m-1}}{x+1}~~ (n:\textrm{odd}) \end{eqnarray} $$ where $\varphi_k = \frac{2k-1}{n} \pi$

11. $$ - x \cos m \varphi + \cos (m-1) \varphi = -(x - \cos \varphi) \cos m \varphi + \sin m \varphi \sin \varphi $$

12.

$$ \int \frac{-(x- \cos \varphi)}{x^2 - 2 x \cos \varphi + 1} dx = \log \left(\frac{1}{\sqrt{x^2 - 2x \cos \varphi + 1}}\right) + C $$

13.

$$ \int \frac{\sin \varphi}{x^2 - 2 x \cos \varphi + 1} dx = \arctan\left(\frac{x- \cos \varphi}{\sin \varphi}\right) + C $$

by introducing new C so that $C \to C + \arctan(\frac{\cos \varphi}{\sin \varphi})$

$$ \int \frac{\sin \varphi}{x^2 - 2 x \cos \varphi + 1} dx = \arctan\left(\frac{\sin \varphi}{1/x - \cos \varphi}\right) + C $$

14. let $A_k(x) = \log \left(\frac{1}{\sqrt{x^2 - 2x \cos \varphi_k +1}}\right)$, $B_k(x)= \arctan \left(\frac{\sin \varphi_k}{1/x - \cos \varphi_k}\right)$ then

$$ \begin{eqnarray} \int \frac{x^{m-1}}{x^n+1} dx &=& \frac{2}{n} \sum_{k=1}^{\frac{n}{2}} \left(\cos m \varphi_k A_k(x) + \sin m \varphi_k B_k(x)\right) ~~ (n:\textrm{even}) \\ \int \frac{x^{m-1}}{x^n+1} dx &=& \frac{2}{n} \sum_{k=1}^{\frac{n-1}{2}} \left(\cos m \varphi_k A_k(x) + \sin m \varphi_k B_k(x)\right) + \frac{(-1)^{m-1}}{n} \log(x+1) ~~ (n:\textrm{odd}) \end{eqnarray} $$

15.

$$ \lim_{x \to \infty} \sum_{k}^{N} \cos m \varphi_k A_k(x) = \lim_{x \to \infty}(- \log x) \sum_k^{N}\cos m \varphi_k $$

16. When x moves from 0 to $\infty$ continuously, $B_k(x)$ must move from 0 to $\pi - \varphi_k$. Not 0 to -$\varphi_k$. Thus

$$ \lim_{x \to \infty} \sum_{k}^{N} \sin m \varphi_k B_k(x) = \sum_{k}^{N} (\pi - \varphi) \sin m \varphi_k $$

17. Let $\omega = \frac{m \pi}{n}$, define S,T,U as follows $$ \begin{eqnarray} S &=& \sum_{k=1}^{N} \cos m \varphi_k = \sum_{k=1}^{N} \cos (2k -1) \omega \\ T &=& \sum_{k=1}^{N} \sin m \varphi_k = \sum_{k=1}^{N} \sin (2k -1) \omega \\ U &=& \sum_{k=1}^{N} \varphi_k \sin m \varphi_k = \frac{\pi}{n} \sum_{k=1}^{N} (2k-1) \sin (2k -1) \omega \end{eqnarray} $$

18. multiply $\sin \omega$ on 17-1,17-2 then it can be shown that

$$ \begin{eqnarray} S &=& \frac{\sin 2 N \omega }{2 \sin \omega} \\ T &=& \frac{1 - \cos 2 N \omega}{2 \sin \omega} \end{eqnarray} $$

19. By differentiating 17-1 with respect to $\omega$,

$$ U = \frac{\pi}{2 n \sin \omega} \left(\frac{\sin (2 N - 1) \omega}{\sin \omega} - (2N - 1) \cos 2 N \omega \right) $$

20.

$$ \lim_{x \to \infty} \frac{2}{n} \sum_{k=1}^{N} \left(\cos m \varphi_k A_k(x) + \sin m \varphi_k B_k(x)\right) \\ = \frac{2}{n} \left(S \lim_{x \to \infty} (- \log x) + \pi T- U\right) $$

21. combining the results from 14,18,19,20, and substituting $2N = n $ for the case of even n, and $2N = n-1 $ for odd n, it can be shown that

$$ \int_0^{\infty} \frac{x^{m-1}}{1+x^n} dx = \frac{\pi}{n \sin (m \pi / n ) } $$