## Monday Math 27

Here’s another definite integral problem: find a general formula for $I_n=\int_0^{\infty}\frac{dx}{1+x^n}$, with n an integer greater than 1.

First, let $u=1+x^n$. Then $du=nx^{n-1}\,dx$, so $dx=\frac{1}{n}(u-1)^{\frac{1}{n}-1}\,du$, and our integral becomes:
$I_n=\frac{1}{n}\int_1^{\infty}u^{-1}(u-1)^{\frac{1}{n}-1}\,du$.
Now, we perform the substitution $t=\frac{1}{u}$, giving:
$\begin{array}{rcl}I_n&=&\frac{1}{n}\int_1^0(t)(\frac{1}{t}-1)^{\frac{1}{n}-1}(-\frac{dt}{t^2})\\&=&\frac{1}{n}\int_0^1t^{-1}(\frac{1}{t}-1)^{\frac{1}{n}-1}\,dt\\&=&\frac{1}{n}\int_0^1t^{-\frac{1}{n}}(1-t)^{\frac{1}{n}-1}\,dt\end{array}$.

We should recoginze this last integral as a beta function:
$\int_0^1t^p(1-t)^q\,dt=B(p+1,q+1)$, so we have:
$\begin{array}{rcl}I_n&=&\frac{1}{n}B(1-\frac{1}{n},\frac{1}{n})\\&=&\frac{1}{n}\frac{\Gamma(1-\frac{1}{n})\Gamma(\frac{1}{n})}{\Gamma(1)}\\&=&\frac{1}{n}\Gamma(1-\frac{1}{n})\Gamma(\frac{1}{n})\end{array}$.
Now, recall the Euler reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{sin(\pi{z})}$. We see that this applies to the above with $z=\frac{1}{n}$, giving us:
$\begin{array}{rcl}I_n&=&\frac{1}{n}\frac{\pi}{sin(\pi/n)}\\&=&\frac{\pi/n}{sin(\pi/n)}\end{array}$.

This is easily confirmed for $I_2=\frac{\pi/2}{sin(\pi/2)}=\frac{\pi}{2}$ by performing the original integral (arctangent). By factoring $x^3+1=(x+1)(x^2-x+1)$ and $x^4+1=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$, and using partial fractions, one can confirm $I_3=\frac{\pi/3}{sin(\pi/3)}=\frac{2\pi}{3\sqrt{3}}$ and $I_4=\frac{\pi/4}{sin(\pi/4)}=\frac{\pi}{2\sqrt{2}}$.
We also see that $I_{\infty}\equiv\lim_{n\to\infty}I_n=1$ as expected by the fact that $\lim_{n\to\infty}\frac{1}{1+x^n}=\begin{cases}1,&{0}\le{x}<1\\{0},&x>1\end{cases}$

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### One Response to “Monday Math 27”

1. Monday Math 76 « Twisted One 151’s Weblog Says:

[…] . How does this relate to our integral? One may recall that . Now, . Thus, . In fact, . (See here for a similar use of the beta function and Euler reflection […]