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 […]

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