How to solve $\int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx$?

Question

As an exercise for myself I constructed the Integral

$$ \int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx $$

with $n\in \mathbb{N}$. With the help of Mathematica I found the analytical result

$$ \int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx=\frac{1+(-1)^n}{4^{n+1}}\Gamma(n+1)\left[\zeta\left(n+1,\frac{1}{4}\right)-\zeta\left(n+1,\frac{3}{4}\right)\right]. $$

For $n=1$ (and probably $n\in \mathbb{N}$) one can employ the methods of complex analysis and find a result. For $n\in \mathbb{N}$ I encountered a nasty recursion relation. I can provide details if needed. Is there another way how to solve the integral at hand?

Answer 1

Referring to this answer, $$I_n=\frac{\pi}2\frac{d^n}{dx^n}\sec\left(\frac{\pi x}2\right)\bigg\vert_{x=0}$$

or equivalently,

$$\int^\infty_0\frac{\log^n(x)}{x^2+1}dx=\left(\frac{\pi}2\right)^{n+1}\sec^{(n)}(0)$$

$$\int^\infty_0\frac{\log^{2n}(x)}{x^2+1}dx=(-1)^nE_{2n}\left( \frac{\pi}2\right)^{2n+1} $$

The integral is zero for odd $n$.

Answer 2

Let, for $n$, a natural integer,

\begin{align}J_n=\int_0^\infty \frac{\ln^n x}{1+x^2}\,dx\end{align}

First, observe that if $n$ is odd then $J_n=0$ (perform the change of variable $y=\dfrac{1}{x}$ )

Consider, for $n$, a natural integer,

\begin{align}K_n=\int_0^\infty \int_0^\infty\frac{\ln^n(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\end{align}

Observe that,

\begin{align}K_{2n}&=\sum_{k=0}^{2n}\binom{2n}{k} J_kJ_{2n-k}\\ &=\sum_{k=0}^{n}\binom{2n}{2k} J_{2k}J_{2(n-k)} \end{align}

On the other hand,

Perform the change of variable $u=xy$ ($x$ variable, $y$, a "parameter"),

\begin{align}K_n&=\int_0^\infty \int_0^\infty\frac{y\ln^n u}{(1+y^2)(u^2+y^2)}\,du\,dy\\ &=\frac{1}{2}\int_0^\infty \left[\ln\left(\frac{1+y^2}{u^2+y^2}\right)\right]_{y=0}^{y=\infty}\frac{\ln^{n} u}{u^2-1}\,du\\ &=\int_0^\infty \frac{\ln^{n+1} u}{u^2-1}\,du\\ &=\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du+\int_1^\infty \frac{\ln^{n+1} u}{u^2-1}\,du\\ \end{align}

In the second integral perform the change of variable $v=\dfrac{1}{u}$,

\begin{align}K_n&=\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du+\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du\\ &=\int_0^1 \frac{\left(1+(-1)^n\right)\ln^{n+1} u}{u^2-1}\,du\\ &=\left(1+(-1)^n\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du-\left(1+(-1)^n\right)\int_0^1\frac{u\ln^{n+1} u}{u^2-1}\,du\\ \end{align}

In the latter integral perform the change of variable $y=u^2$,

\begin{align}K_n&=\left(1+(-1)^n\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du-\frac{1+(-1)^n}{2^{n+2}}\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du\\ &=\left(1+(-1)^n\right)\left(1-\frac{1}{2^{n+2}}\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du\\ &=\left(1+(-1)^n\right)\left(1-\frac{1}{2^{n+2}}\right)(-1)^{n+2}(n+1)!\zeta(n+2) \end{align}

Therefore, one obtains a recursion relation,

\begin{align} \boxed{\sum_{k=0}^{n}\binom{2n}{2k} J_{2k}J_{2(n-k)}=2\left(1-\frac{1}{2^{2n+2}}\right)(2n+1)!\zeta(2n+2)} \end{align}

Observe that,

\begin{align}J_0=\frac{\pi}{2}\end{align}

Answer 3

As you have said, this problem can be solved by contour integration, and in my opinion, the solution is easier than the other methods described in this answer.

Starting with some basic manipulation, we have: $$ \int_0^\infty \frac{\ln^n x}{x^2+1} \,dx \stackrel{u=\ln x}{=} \int_{-\infty}^\infty \frac{u^n e^u}{e^{2u}+1} \,du = \int_{-\infty}^\infty \frac{u^n}{2 \cosh{u}} \,du $$ Let $$ f(z)=\frac{z^n}{2 \cosh{z}}. $$ $f$ has only simple poles at $\displaystyle \left\{\left.\frac{(2k+1)\pi i}{2} ~\right|~ k \in \mathbb{Z}\right\}$, and we have $$ \mathop{\text{Res}}_{z={(2k+1)\pi i / 2}} f(z) = \frac{(-1)^k(2k+1)^n \pi^n i^{n-1}}{2^{n+1}} $$ Let $\Gamma_R$ be the semicircle in the upper half-plane of radius $R$, traversed counterclockwise. By the estimation lemma, we have: $$ \left|\int_{\text{Arc}} f(z) \,dz\right| \le \pi\sup_{\theta \in [0, \pi]}\left|\frac{R^n e^{i n \theta}}{\cosh{Re^{i \theta}}}\right| \to 0 \text{ as } R \to \infty $$ So as $R \to \infty$, we're left with: $$ \begin{align*} \int_{-\infty}^\infty f(z)\,dz & = \frac{\pi^{n+1} i^n}{2^n} \sum_{k=0}^\infty (-1)^k(2k+1)^n \\ & = \frac{\pi^{n+1} i^n}{2^n} \beta(-n). \\ \end{align*} $$ where $\beta(s)$ is the Dirichlet beta function. $\beta(-n)=0$ for all odd $n$, so let's consider even $n$ specifically. Let $n=2m$, and we have: $$ \begin{align*} \int_0^\infty \frac{\ln^{2m} x}{x^2+1} \,dx & = \frac{(-1)^m\pi^{2m+1}}{2^{2m}}\beta(-2m) \\ & = \frac{(-1)^m\pi^{2m+1}}{2^{2m+1}}E_{2m} \end{align*} $$ where $E_{m}$ is the $m$th Euler number. I'll leave it up to the reader to show that the result here and the expression in the question are equivalent.

Answer 4

We start how @mediocrevegetable1 begins their answer: $$I_n=\int_{-\infty}^\infty\frac{u^ne^u}{1+e^{2u}}du$$ Split this integral into two parts and rewrite: $$I_n=(1-(-1)^n)\int_0^\infty\frac{u^ne^{-u}}{1+e^{-2u}}du$$So we only consider $$I_{2n}=2\int_0^\infty\frac{u^{2n}e^{-u}}{1+e^{-2u}}du$$Using the geometric series formula, where the common ratio is $-e^{-2u}$, we get $$I_{2n}=2\sum_{k=0}^\infty(-1)^k\int_0^\infty u^{2n}e^{-(2k+1)u}du$$Let $v=(2k+1)u$ and then apply the integral definition of the Gamma function: $$I_{2n}=2\sum_{k=0}^\infty\frac{(-1)^k\Gamma(2n+1)}{(2k+1)^{2n+1}}=2\Gamma(2n+1)\beta(2n+1)$$Where $\beta$ is the dirichlet beta function.