Do we have a closed form for $\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t $?

Question

Latest Edit

We are glad to see there are 4 alternative solutions which give the same closed form to the integral:

$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a^2+t^2\right)^n} d t = \frac{a^{1-2n}\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] },$$

where $\psi$ denotes the Digamma Function.

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In the post, we found that $$\int_0^{\infty} \frac{\ln x}{a^2+t^2} d x =\frac{\pi \ln a}{2 a }$$ Now I want to generalise the integral as $$ I_n=\int_0^{+\infty} \frac{\ln t}{\left(a+t^2\right)^n} d t $$ where $n\in N$.

Replacing a by $\sqrt{a}$ and differentiating both sides w.r.t. $a$ by $n$ times yields $$ \begin{aligned} &J(a)=\int_0^{\infty} \frac{\ln t}{a+t^2} d t=\frac{\pi}{4 \sqrt{a}} \ln a\\ &\frac{d^n}{d a^n}(J(a))=\frac{\pi}{4} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) \\ & (-1)^n n ! \int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)} d t= \frac{\pi}{4} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)\\& \boxed{\int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)^{n+1}} d t=\frac{(-1)^n \pi}{4 n !} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)} \end{aligned} $$ By Wolfram-alpha, we have

$$ \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)=(-1)^n a^{-\frac{1}{2}-n}\left(\frac{1}{2}\right)_n\left(\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)\cdots (*) $$ Hence

$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)^{n+1}} d t=\frac{\pi}{4 n !} a^{-\frac{1}{2}-n}\left(\frac{1}{2}\right)_n\left(\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)}$$

In particular, when $a=1$, we have $$\boxed{\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^{n+1}} d t=\frac{\pi}{4 n !} \left(\frac{1}{2}\right)_n\left(\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)}$$ For example, $$$$ \begin{aligned} \int_0^\infty \frac{\ln t}{\left(1+t^2\right)^4} d t=& \frac{\pi}{24} \cdot \frac{15}{8}\left(-\gamma -\ln 4-\frac{46}{15}+\gamma +\ln 4\right) =-\frac{23 \pi}{96} \end{aligned} $$ $$ My Question: How to find a closed form for $\frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) $?

Answer 1

The generalized Leibniz rule gives

$$\newcommand{\d}{\mathrm{d}} \frac{\d^n}{\d x^n} \ln(x) x^{-1/2} \bigg|_{x=1} = \sum_{k=0}^n \binom n k \left( \frac{\d^{n-k}}{\d x^{n-k}} \ln(x) \bigg|_{x=1} \right)\left( \frac{\d ^{k}}{\d x^{k}} x^{-1/2} \bigg|_{x=1} \right)$$

For the $(n-k)$th derivative of $\ln(x)$, it is easy to show by, e.g., induction that

$$\frac{\d ^{n-k}}{\d x^{n-k}} \ln(x) = \frac{(n-k-1)! \cdot (-1)^{n-k-1}}{x^{n-k}}$$

Similarly,

$$\begin{align*} \frac{\d ^k}{\d x^k} x^{-1/2} &= \frac{-1}{2} \frac{-3}{2} \frac{-5}{2} \cdots \frac{ -(2k-1)}{2} x^{-k-1/2} \\ &= \frac{(-1)^k (2k-1)!!}{2^k} x^{-k-1/2} \end{align*}$$

so

$$\begin{align*} \frac{\d ^n}{\d x^n} \ln(x) x^{-1/2} \bigg|_{x=1} &= \sum_{k=0}^n \binom n k (n-k-1)! \cdot (-1)^{n-k-1} \cdot \frac{(-1)^k (2k-1)!!}{2^k} \\ &= (-1)^{n-1} \cdot \sum_{k=0}^n \binom n k (n-k-1)! \cdot \frac{(2k-1)!!}{2^k} \\ &= (-1)^{n-1} n! \cdot \sum_{k=0}^n \frac{1}{(n-k) \cdot k!} \cdot \frac{(2k-1)!!}{2^k} \\ \end{align*}$$

I don't know if there's a meaningful simplification beyond this.

Answer 2

Assuming that $n$ is a positive integer, using hypergeometric functions$$I_n=\int \frac{\log(t)}{\left(1+t^2\right)^n}\, dt=t \log (t) \, _2F_1\left(\frac{1}{2},n;\frac{3}{2};-t^2\right)-t \, _3F_2\left(\frac{1}{2},\frac{1}{2},n;\frac{3}{2},\frac{3}{2};-t^2 \right)$$ $$J_n=\int_0^\infty \frac{\log(t)}{\left(1+t^2\right)^n}\, dt=-\frac{\sqrt{\pi }}{4}\,\frac{\Gamma \left(n-\frac{1}{2}\right)}{\Gamma (n)}\,\left(H_{n-\frac{3}{2}}+2 \log (2)\right)$$

Asymptotically $$J_n=-\frac{\sqrt{\pi }}{4 \sqrt{n}}(\log (n)+\gamma +2 \log (2))+O\left(\frac{1}{n^{3/2}}\right)$$ Trying to simplify what is given in other answers $$\frac{d^n}{d a^n}\left(\frac{\log( a)}{\sqrt{a}}\right)=\sqrt \pi \frac{ \left(\log (a)-H_{n-\frac{1}{2}}-2 \log (2)\right)}{a^{n+\frac{1}{2}}\,\,\Gamma \left(\frac{1}{2}-n\right)}$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} {\tt I}_{n} & \equiv\color{#44f}{\int_{0}^{\infty} {\ln\pars{t} \over \pars{a + t^{2}}^{n}}\,\dd t} \sr{t^{2}\ \mapsto\ t}{=} \int_{0}^{\infty}{\ln\pars{t}/2 \over \pars{a + t}^{n}}\pars{{1 \over 2}\,t^{-1/2}}\,\dd t \\[5mm] & = \left.{1 \over 4}\partiald{}{\nu}\int_{0}^{\infty}{\,t^{\pars{\nu + 1/2} - 1 \,\,\,} \over \pars{a + t}^{n}}\,\dd t\right\vert_{\nu\ =\ 0} \end{align} Note that \begin{align} & \pars{a + t}^{-n} = a^{-n}\sum_{k = 0}^{\infty}{-n \choose k}\pars{t \over a}^{k} = a^{-n}\sum_{k = 0}^{\infty}{n + k - 1 \choose k} {\pars{-t}^{k} \over a^{k}} \\[5mm] = & \ \sum_{k = 0}^{\infty} {a^{-n - k} \over \pars{n - 1}!}\,\Gamma\pars{n + k}{\pars{-t}^{k} \over k!} \end{align} Therefore, \begin{align} {\tt I}_{n} & \equiv\color{#44f}{\int_{0}^{\infty} {\ln\pars{t} \over \pars{a + t^{2}}^{n}}\,\dd t}\qquad \substack{\mbox{With}\\ \ds{Ramanujan's\ Master\ Theorem}}\ : \\[5mm] & = {1 \over 4}\partiald{}{\nu}\bracks{\Gamma\pars{\nu + {1 \over 2}} {a^{-n + \nu + 1/2}\,\,\, \over \pars{n - 1}!}\,\Gamma\pars{n - \nu - {1 \over 2}}}_{\nu\ =\ 0} \\[5mm] & = {\root{\pi}a^{-n + 1/2} \over 4\pars{n - 1}!}\Gamma\pars{n - {1 \over 2}} \bracks{\ln\pars{a} + \Psi\pars{1 \over 2} - \Psi\pars{n - {1 \over 2}}} \\[5mm] & = \color{#44f}{{\root{\pi}a^{-n + 1/2}\,\, \over 4\pars{n - 1}!}\,\Gamma\pars{n - {1 \over 2}}} \times \\ & \rule{1cm}{0pt}\color{#44f}{\bracks{\ln\pars{a} - \gamma - 2\ln\pars{2} - \Psi\pars{n - {1 \over 2}}}} \\ & \end{align} $\ds{\gamma:\ Euler\mbox{-}mascheroni\ Constant.\ \Psi:\ Digamma\ Function}$. $\ds{\Psi\pars{n - 1/2}}$ can be evaluated with the $\ds{Gauss\ Digamma\ Theorem}$ for given values of $\ds{n \in \mathbb{N}_{\geq\ 1}}$.

Answer 4

As @metamorphy suggested, let $I\left(\lambda\right)=\int_0^{\infty} \frac{t^{2 \lambda-1}}{\left(a^2+t^2\right)^n} d t,$ then $$I_n=\int_0^{\infty} \frac{\ln t}{\left(a^2 +t^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) .$$ Now we are going to express $I\left(\lambda\right)$ as a beta function by letting $t=a\tan \theta$, then $$ \begin{aligned} I(\lambda) &= a^{2(\lambda-n)}\int_0^{\frac{\pi}{2}} \sin ^{2 \lambda-1} \theta \cos ^{2(n-\lambda)-1} \theta d \theta \\ &=\frac{a^{2(\lambda-n)}}{2} B(\lambda, n-\lambda) \\ &=\frac{a^{2(\lambda-n)}}{2\Gamma(n) } \Gamma(\lambda) \Gamma(n-\lambda) \end{aligned} $$ By logarithmic differentiation, we get $$ \begin{aligned} &\frac{I^{\prime}(\lambda)}{I(\lambda)}=2\ln a+\psi(\lambda)-\psi(n-\lambda) \\ &I^{\prime}(\lambda)=\frac{a^{2(\lambda-n)}}{2 \Gamma(n)} \Gamma(\lambda) \Gamma(n-\lambda)[ 2\ln a+\psi(\lambda)-\psi(n-\lambda)] \end{aligned} $$ Putting $\lambda=\frac{1}{2} $ yields $$ I^{\prime}\left(\frac{1}{2}\right)=\frac{a^{1-2n}\sqrt{\pi}}{2} \cdot \frac{\Gamma\left(n-\frac{1}{2}\right)}{\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] $$ Hence $$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a^2+t^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) = \frac{a^{1-2n}\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] },$$

which is same as the answer provided by @Felix Marin.

Back to our integral,

$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) = \frac{\sqrt{\pi}\Gamma \left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] }$$

By Wolfram-Alpha, $\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right) = -H_{n-\frac{3}{2}}-2 \log (2) $ yields $$I_n= -\frac{\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left(H_{n-\frac{3}{2}}+2\ln 2\right),$$ which is the same as the answer provided by @Claude Leibovici.

Answer 5

Note $$ \int_0^\infty\frac{t^a}{(1+t)^n}dt=\int_0^1u^a(1-u)^{n-a-2}du=B(a+1,n-a-1)=\frac{1}{(n-1)!}\Gamma(a+1)\Gamma(n-a-1) $$ and hence $$ \int_0^\infty\frac{t^a\ln t}{(1+t)^n}dt=\frac{d}{da}\frac{\Gamma(a+1)\Gamma(n-a-1)}{\Gamma(n)}=\frac{1}{(n-1)!}\Gamma (a+1) \Gamma (-a+n-1) (\psi ^{(0)}(a+1)-\psi ^{(0)}(-a+n-1)). $$ So \begin{eqnarray} I_n&=&\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t\\ &=&\frac14\int_0^{\infty} \frac{t^{-1/2}\ln t}{\left(1+t\right)^n} d t \\ &=&\frac14\frac{1}{(n-1)!}\Gamma (a+1) \Gamma (-a+n-1) (\psi ^{(0)}(a+1)-\psi ^{(0)}(-a+n-1))\bigg|_{a=-\frac12} \\ &=&\frac{\sqrt{\pi }\Gamma \left(n-\frac{1}{2}\right)}{4(n-1)!}\bigg[ \psi ^{(0)}\left(\frac{1}{2}\right) -\psi ^{(0)}\left(n-\frac{1}{2}\right)\bigg]. \end{eqnarray}