Calculus – Integration – A remarkable class of integrals

We will prove that \(\displaystyle\int_0^{\infty}\frac{dx}{(1+x^2)(1+x^k)}=\frac{1}{4}\pi\) for all \(k\in\mathbb{Z}=\{\ldots,-2,-1,0,1,2,\ldots\}\).

For \(k=0\) this is easy: \(\displaystyle\int_0^{\infty}\frac{dx}{(1+x^2)(1+x^0)}=\frac{1}{2}\int_0^{\infty}\frac{dx}{1+x^2}=\frac{1}{2}\arctan(x)\bigg|_0^{\infty}=\frac{1}{4}\pi\).

For \(k\in\{1,2,3,\ldots\}\) we note that

\[\int_0^{\infty}\frac{dx}{(1+x^2)(1+x^k)}=\int_0^1\frac{dx}{(1+x^2)(1+x^k)}+\int_1^{\infty}\frac{dx}{(1+x^2)(1+x^k)}.\]

Using \(x=1/t\) we find that

\[\int_1^{\infty}\frac{dx}{(1+x^2)(1+x^k)}=\int_1^0\frac{1}{(1+t^{-2})(1+t^{-k})}\left(-\frac{1}{t^2}\right)\,dt =\int_0^1\frac{dt}{(t^2+1)(1+t^{-k})}.\]

So we conclude that for \(k\in\{1,2,3,\ldots\}\):

\begin{align*} \int_0^{\infty}\frac{dx}{(1+x^2)(1+x^k)}&=\int_0^1\frac{dx}{(1+x^2)(1+x^k)}+\int_0^1\frac{dx}{(1+x^2)(1+x^{-k})}\\[2.5mm] &=\int_0^1\frac{1}{1+x^2}\left(\frac{1}{1+x^k}+\frac{1}{1+x^{-k}}\right)\,dx=\int_0^1\frac{1}{1+x^2}\left(\frac{1}{1+x^k}+\frac{x^k}{x^k+1}\right)\,dx\\[2.5mm] &=\int_0^1\frac{1}{1+x^2}\cdot\frac{1+x^k}{1+x^k}\,dx=\int_0^1\frac{dx}{1+x^2}=\arctan(x)\bigg|_0^1=\frac{1}{4}\pi. \end{align*}

Similarly we consider

\[\int_0^{\infty}\frac{dx}{(1+x^2)(1+x^{-k})}=\int_0^1\frac{dx}{(1+x^2)(1+x^{-k})}+\int_1^{\infty}\frac{dx}{(1+x^2)(1+x^{-k})}\]

for \(k\in\{1,2,3,\ldots\}\). Again we use \(x=1/t\) to find that

\[\int_1^{\infty}\frac{dx}{(1+x^2)(1+x^{-k})}=\int_1^0\frac{1}{(1+t^{-2})(1+t^k)}\left(-\frac{1}{t^2}\right)\,dt =\int_0^1\frac{dt}{(t^2+1)(1+t^k)}.\]

Hence we have for \(k\in\{1,2,3,\ldots\}\)

\begin{align*} \int_0^{\infty}\frac{dx}{(1+x^2)(1+x^{-k})}&=\int_0^1\frac{dx}{(1+x^2)(1+x^{-k})}+\int_0^1\frac{dx}{(1+x^2)(1+x^k)}\\[2.5mm] &=\int_0^1\frac{1}{1+x^2}\left(\frac{1}{1+x^{-k}}+\frac{1}{1+x^k}\right)\,dx=\int_0^1\frac{1}{1+x^2}\left(\frac{x^k}{x^k+1}+\frac{1}{1+x^k}\right)\,dx\\[2.5mm] &=\int_0^1\frac{1}{1+x^2}\cdot\frac{x^k+1}{x^k+1}\,dx=\int_0^1\frac{dx}{1+x^2}=\arctan(x)\bigg|_0^1=\frac{1}{4}\pi. \end{align*}

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Last modified on March 8, 2021