Let $n \in \mathbb{N}$. Find a closed form for the integral:
$$\mathcal{J}=\int_0^\infty \frac{\log^n x}{1+x^2}\, {\rm d}x$$
Solution
We have successively:
\begin{align*} \int_{0}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x + \int_{1}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x - \int_{1}^{0} \frac{\log^n \left ( \frac{1}{x} \right )}{1+ \left ( \frac{1}{x} \right )^2} \frac{1}{x^2} \, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2} \, {\rm d}x + \int_{0}^{1} \frac{\log^n \left ( \frac{1}{x} \right )}{1+x^2}\, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x+ \int_{0}^{1} \frac{(-1)^n \log^n x}{1+x^2} \, {\rm d}x \\ &= \left ( 1+(-1)^n \right ) \int_{0}^{1} \frac{\log^n x}{1+x^2} \, {\rm d}x\\ &=\left ( 1+(-1)^n \right )\int_{0}^{1} \log^n x \sum_{k=0}^{\infty} (-1)^k x^{2k} \, {\rm d}x \\ &= \left ( 1+(-1)^n \right ) \sum_{k=0}^{\infty} (-1)^k \int_{0}^{1}x^{2k} \log^n x \, {\rm d}x\\ &= \left ( 1+(-1)^n \right )(-1)^n n! \sum_{k=0}^{\infty} \frac{(-1)^k}{\left ( 2k+1 \right )^{n+1}} \\ &= \left ( 1+(-1)^n \right )(-1)^n n! \sum_{k=0}^{\infty} \left [ \frac{1}{\left ( 4k+1 \right )^{n+1}} - \frac{1}{\left ( 4k+3 \right )^{n+1}} \right ] \\ &=\left ( 1+(-1)^n \right )(-1)^n n! \frac{1}{4^{n+1}} \sum_{k=0}^{\infty} \left [ \frac{1}{\left ( k+ \frac{1}{4} \right )^{n+1}} - \frac{1}{\left ( k+\frac{3}{4} \right )^{n+1}} \right ] \\ &= \frac{1}{4^{n+1}}\left ( 1+(-1)^n \right ) \left [ \psi^{(n)} \left ( \frac{3}{4} \right ) - \psi^{(n)} \left ( \frac{1}{4} \right )\right ] \end{align*}
Distinguishing cases for $n$ (either $n$ is odd or even) we obtain the beautiful formula
$$ \int_{0}^{\infty}\frac{\log^n x}{1+x^2}\, {\rm d}x = \left\{\begin{matrix} 0 &, &n \; {\rm odd} \\\\ \displaystyle \frac{2}{4^{n+1}}\left [ \psi^{(n)} \left ( \frac{3}{4} \right ) - \psi^{(n)} \left ( \frac{1}{4} \right ) \right ]& ,& n \; {\rm even} \end{matrix}\right.$$
However, if $n$ is even from the polygamma reflection formula
$$ \psi^{(n)}(1-z)-\psi^{(n)} (z)= \pi \; \frac{\mathrm{d}^{n} }{\mathrm{d} x^{n}} \cot \pi z$$
thus if $n$ is even we have (the non so beautiful form):
$$ \int_{0}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x = \left[ \frac{2\pi}{4^{n+1}} \frac{\mathrm{d}^n }{\mathrm{d} x^n} \cot \pi z \right]_{z=1/4} \;\; n \; \; {\rm even}$$
$$\mathcal{J}=\int_0^\infty \frac{\log^n x}{1+x^2}\, {\rm d}x$$
Solution
We have successively:
\begin{align*} \int_{0}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x + \int_{1}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x - \int_{1}^{0} \frac{\log^n \left ( \frac{1}{x} \right )}{1+ \left ( \frac{1}{x} \right )^2} \frac{1}{x^2} \, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2} \, {\rm d}x + \int_{0}^{1} \frac{\log^n \left ( \frac{1}{x} \right )}{1+x^2}\, {\rm d}x\\ &= \int_{0}^{1} \frac{\log^n x}{1+x^2}\, {\rm d}x+ \int_{0}^{1} \frac{(-1)^n \log^n x}{1+x^2} \, {\rm d}x \\ &= \left ( 1+(-1)^n \right ) \int_{0}^{1} \frac{\log^n x}{1+x^2} \, {\rm d}x\\ &=\left ( 1+(-1)^n \right )\int_{0}^{1} \log^n x \sum_{k=0}^{\infty} (-1)^k x^{2k} \, {\rm d}x \\ &= \left ( 1+(-1)^n \right ) \sum_{k=0}^{\infty} (-1)^k \int_{0}^{1}x^{2k} \log^n x \, {\rm d}x\\ &= \left ( 1+(-1)^n \right )(-1)^n n! \sum_{k=0}^{\infty} \frac{(-1)^k}{\left ( 2k+1 \right )^{n+1}} \\ &= \left ( 1+(-1)^n \right )(-1)^n n! \sum_{k=0}^{\infty} \left [ \frac{1}{\left ( 4k+1 \right )^{n+1}} - \frac{1}{\left ( 4k+3 \right )^{n+1}} \right ] \\ &=\left ( 1+(-1)^n \right )(-1)^n n! \frac{1}{4^{n+1}} \sum_{k=0}^{\infty} \left [ \frac{1}{\left ( k+ \frac{1}{4} \right )^{n+1}} - \frac{1}{\left ( k+\frac{3}{4} \right )^{n+1}} \right ] \\ &= \frac{1}{4^{n+1}}\left ( 1+(-1)^n \right ) \left [ \psi^{(n)} \left ( \frac{3}{4} \right ) - \psi^{(n)} \left ( \frac{1}{4} \right )\right ] \end{align*}
Distinguishing cases for $n$ (either $n$ is odd or even) we obtain the beautiful formula
$$ \int_{0}^{\infty}\frac{\log^n x}{1+x^2}\, {\rm d}x = \left\{\begin{matrix} 0 &, &n \; {\rm odd} \\\\ \displaystyle \frac{2}{4^{n+1}}\left [ \psi^{(n)} \left ( \frac{3}{4} \right ) - \psi^{(n)} \left ( \frac{1}{4} \right ) \right ]& ,& n \; {\rm even} \end{matrix}\right.$$
However, if $n$ is even from the polygamma reflection formula
$$ \psi^{(n)}(1-z)-\psi^{(n)} (z)= \pi \; \frac{\mathrm{d}^{n} }{\mathrm{d} x^{n}} \cot \pi z$$
thus if $n$ is even we have (the non so beautiful form):
$$ \int_{0}^{\infty} \frac{\log^n x}{1+x^2}\, {\rm d}x = \left[ \frac{2\pi}{4^{n+1}} \frac{\mathrm{d}^n }{\mathrm{d} x^n} \cot \pi z \right]_{z=1/4} \;\; n \; \; {\rm even}$$
A relative integral is the following:
ReplyDelete$$\int_0^\infty \frac{\log x}{(1+x^2)^m}\, {\rm d}x, \; m \in \mathbb{N}$$
Solution
Let us begin with the following two known represantations of the Beta function:
$${\rm B}(x, y)= \int_{0}^{\infty} \frac{t^{x-1}}{\left ( 1+t \right )^{x+y}}\, {\rm d}t = \frac{\Gamma (x)\Gamma (y)}{\Gamma \left ( x+y \right )}$$
Now, setting $t \mapsto t^2$ as well as $y \mapsto m -x$ we have that:
\begin{equation} {\rm B} \left ( x, m-x \right )= m \int_{0}^{\infty} \frac{t^{2x-1}}{\left ( 1+t^2 \right )^m}\, {\rm d}t \end{equation}
Now, differentiating $(1)$ with respect to $x$ once we have that:
$$\int_{0}^{\infty} \frac{t^{2x-1} \log x }{\left ( 1+t^2 \right )^m}\, {\rm d}t = \frac{\Gamma (x)\Gamma(m-x) \bigg [ \psi^{(0)}(x)- \psi^{(0)}(m-x) \bigg ]}{m^2 \Gamma(m)}$$
Thus our required integral is equal to:
$$ \int_{0}^{\infty} \frac{\log x}{\left ( 1+t^2 \right )^m}\, {\rm d}x ={\rm B}^{(1)} \left ( \frac{1}{2}, m- \frac{1}{2} \right ) = \frac{\Gamma \left ( \frac{1}{2} \right ) \Gamma \left ( m- \frac{1}{2} \right ) \bigg[ \psi^{(0)}\left ( \frac{1}{2}\right) - \psi^{(0)} \left ( m- \frac{1}{2} \right ) \bigg]}{m^2 \Gamma(m)}$$