How do we find the exact value of $\int_{0}^{\infty} \frac{\ln ^{n}\left(1+x^{2}\right)}{1+x^{2}} d x$, where $n\in \mathbb N?$

Question

Latest Edit

By @KStarGamer’s help, I can finally find a reduction formula for $I_n$ as below: $$\boxed{I_n= 2 \ln 2 I_{n-1}+ (n-1)!\sum_{k=0}^{n-2} \frac{2^{n-k}-2}{k!}\zeta(n-k) I_k} $$

where $n\geq 2.$

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In my post, I started to evaluate $$I_1=\int_{0}^{\infty} \frac{\ln \left(1+x^{2}\right)}{1+x^{2}} d x =\pi \ln 2, $$ then I challenge myself on $$I_2=\int_{0}^{\infty} \frac{\ln ^{2}\left(1+x^{2}\right)}{1+x^{2}}dx$$ Again, letting $x=\tan \theta$ as for $I_1$ yields$$I_2=\int_{0}^{\frac{\pi}{2}} \ln ^{2}\left(\sec ^{2} \theta\right) d \theta= 4 \int_{0}^{\frac{\pi}{2}} \ln ^{2}(\cos x) dx $$ It’s very hard to deal with $\ln^2$ and I was stuck. Suddenly a wonderful identity came to my mind. $$ 2\left(a^{2}+b^{2}\right)=(a+b)^{2}+(a-b)^{2}, \\$$

by which

$\displaystyle 2\left[\ln ^{2}(\sin x)+\ln ^{2}(\cos x)\right]=[\ln (\sin x)+\ln (\cos x)]^{2}+[\ln (\sin x)-\ln (\cos x)]^{2} ,\tag*{}\\ $

we have $\displaystyle 4 L=\underbrace{\int_{0}^{\frac{\pi}{2}} \ln ^{2}\left(\frac{\sin 2 x}{2}\right)}_{J} d x+\underbrace{\int_{0}^{\frac{\pi}{2}} \ln ^{2}(\tan x) d x}_{K} \tag*{}\\ $

For the first integral, using $ \int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x=-\dfrac{\pi}{2} \ln 2 $ yields $ \begin{aligned}J &=\int_{0}^{\frac{\pi}{2}}[\ln (\sin 2 x)-\ln 2]^{2} d x \\&=\int_{0}^{\frac{\pi}{2}} \ln ^{2}(\sin 2 x) d x-2 \ln 2 \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x) d x +\frac{\pi \ln ^{2} 2}{2} \\& \stackrel{x\mapsto 2x}{=} \frac{1}{2} \int_{0}^{\pi} \ln ^{2}(\sin x) d x-\ln 2 \int_{0}^{\pi} \ln (\sin x) d x+\frac{\pi \ln ^{2} 2}{2} \\& \stackrel{symmetry}{=} L-\ln 2(-\pi \ln 2)+\frac{\pi \ln ^{2} 2}{2} \\&=L+\frac{3 \pi \ln ^{2} 2}{2}\end{aligned}\tag*{} \\$

For the second integral, letting $ y=\tan x $ and using my post yields

$ \displaystyle K=\int_{0}^{\infty} \frac{\ln ^{2} y}{1+y^{2}} d y=\frac{\pi^{3}}{8}, \tag*{} \\$

then

$ \displaystyle 4L=L+\frac{3 \pi \ln ^{2} 2}{2}+\frac{\pi^{3}}{8} \Rightarrow L=\frac{\pi^{3}}{24}+\frac{\pi \ln ^{2} 2}{2}\tag*{} $ Hence $ \displaystyle \boxed{I_2=4L= \frac{\pi^{3}}{6}+2 \pi \ln ^{2} 2} \tag*{} $

My Question: Can I go further with $I_n$, where $n\geq 3$?

Answer 1

With differentiation of $$\frac2\pi\int_{0}^{\frac\pi2} (2\cos t)^adt = \frac{\Gamma(a+1)}{\Gamma^2\left(\frac a2+1\right)}$$ in $a$, it can be shown that $J_k=\frac2\pi \int_{0}^{\frac\pi2}\ln^k(2\cos t)dt$ are the power-series coefficients of the exponential function below $$ \exp\bigg( \sum_{k=1}^\infty \frac{(-1)^k}k (1-2^{1-k})\zeta(k)x^k\bigg)= \sum_{k=0}^\infty \frac{J_k}{k!}x^k $$ which results in \begin{align} J_0 &= 1,\>\>\>\>\> J_1= 0, \>\>\>\>\> J_2=\frac12 \zeta(2),\>\>\>\>\> J_3= -\frac32 \zeta(3),\>\>\>\>\> J_4= \frac{57}{8} \zeta(4) \\ J_5 &= -\frac{15}2\zeta(2)\zeta(3) -\frac{45}2 \zeta(5),\>\>\>\>\> J_6= \frac{45}2 \zeta^2(3)+ \frac{12375}{64} \zeta(6),\>\>\>\>\> J_7= \cdots \end{align}

Now, note that \begin{align} I_n & = \int_{0}^{\infty} \frac{\ln ^{n}(1+x^{2})}{1+x^{2}} d x\\ &= \int_{0}^{\frac{\pi}{2}} \ln ^{n}\sec ^{2} t\> d t= 2^{n-1}\pi\sum_{k=0}^n \binom nk (-1)^{k} \ln^{n-k}(2)\>J_{k} \end{align} Thus \begin{align} I_1 &= \pi\ln2 \\ I_2 & = \pi [\zeta(2) +2\ln^22]\\ I_3 &=\pi [6\zeta(3) +6\ln2\> \zeta(2) +4\ln^32] \\ I_4 &= \pi[57\zeta(4)+48\ln2\>\zeta(3)+24\ln^22\>\zeta(2)+8\ln^42]\\ I_5 &= \pi[120\zeta(2)\zeta(3)+360\zeta(5)+570\ln2\>\zeta(4)\\ &\hspace{2cm}+240\ln^22\>\zeta(3)+80\ln^32\>\zeta(2)+16\ln^52]\\ I_6&=\cdots \end{align}

Answer 2

Taking inspiration from the answer to this similar question.

$$\int_{0}^{\infty} (1+x^2)^s \, dx = \frac{\sqrt{\pi} \, \Gamma \left(-\frac{2s+1}{2}\right)}{2 \Gamma \left(-s\right)}$$

Valid for $\Re (s) < -\frac{1}{2}$.

Differentiating twice with respect to $s$ and taking the limit as $s \to -1$ gives the desired result. Differentiating once also gives the answer to your previous question.

Differentiate $n$ times to determine $I_n$.

Answer 3

Inspired by KStarGamer that $$ I(a):=\int_{0}^{\infty}\left(1+x^{2}\right)^{a} d x \Rightarrow \quad I^{\prime}(a)=\int_{0}^{\infty}\left(1+x^{2}\right)^{a} \ln \left(1+x^{2}\right) d x $$ Consequently $$ \boxed{I_{n}=I^{(n)}(-1)}, $$ which switches the integration problem to a differentiation problem.

I first express $I(a)$ in terms of Gamma functions as KStarGamer said. $$ \begin{aligned} I(a)&\stackrel{x=\tan \theta}{=} \int_{0}^{\frac{\pi}{2}} \sin ^{2\left(\frac{1}{2}\right)-1} \theta \cos ^{2\left(-a-\frac{1}{2}\right)-1} \theta d \theta\\& =\frac{1}{2} B\left(-a-\frac{1}{2}, \frac{1}{2}\right)\\&=\frac{\sqrt{\pi} \Gamma\left(-a-\frac{1}{2}\right)}{2 \Gamma(-a)}, \end{aligned}$$

By logarithmic differentiation,$$\frac{I^{\prime}(a)}{I(a)}=-\psi\left(-a-\frac{1}{2}\right)+\psi(-a)$$ $$ \boxed{I^{\prime}(a)= I(a)\left[ \psi(-a)-\psi\left(-a-\frac{1}{2}\right)\right]\cdots (*)} $$ where $\psi(z)=\frac{d}{d z}[\ln \Gamma(z)]$.

Using Leibniz Rule, differentiating both sides in (*) w.r.t. $a$ by $n-1$ times followed by putting $a=-1$ yields $$ I^{(n)}(-1)=\sum_{k=0}^{n-1} (-1)^{n-k-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) I^{(k)}(-1)\left[\psi^{(n-k-1)}(1)-\psi^{(n-k-1)}\left(\frac{1}{2}\right)\right] $$ which is a reduction formula involving Polygamma function $\psi(\cdot).$

Converting the formula back to our integral $I_n=I^{(n)}(-1) $, $$ \boxed{I_n=\sum_{k=0}^{n-1} (-1)^{n-k-1}\left(\begin{array}{c} n-1 \\ k \end{array}\right) I_k\left[\psi^{(n-k-1)}(1)-\psi^{(n-k-1)}\left(\frac{1}{2}\right)\right]} $$

Let’s try with the initial $I_0=I^{(0)}(-1)=\frac{\pi}{2}$, applying (*) yields $$ I_{1}=I_0\left[\psi(1)-\psi\left(\frac{1}{2}\right)\right]=\frac{\pi}{2} \ln 4=\pi \ln 2 $$ $$ \begin{aligned} I_2=&-\left(\begin{array}{l} 1 \\ 0 \end{array}\right) I_0\left[\psi^{\prime}(1)-\psi^{\prime}\left(\frac{1}{2}\right)\right] +\left(\begin{array}{l} 1 \\ 1 \end{array}\right) I_1\left[\psi(1)-\psi\left(\frac{1}{2}\right)\right] \\ \end{aligned} $$ Hence $$\int_{0}^{\infty} \frac{\ln ^2\left(1+x^{2}\right)}{1+x^{2}} d x= -\frac{\pi}{2}\left(-\frac{\pi^{2}}{3}\right)+\pi \ln 2 \ln 4= \frac{\pi^{3}}{6}+2 \pi \ln ^{2} 2$$ Furthermore $$ \begin{aligned} \int_{0}^{\infty} \frac{\ln ^{3}\left(1+x^{2}\right)}{1+x^{2}} =&\left(\begin{array}{l} 2 \\ 0 \end{array}\right) I_0\left[\psi^{(2)}(1)-\psi^{(2)}\left(\frac{1}{2}\right)\right] \\ &-\left(\begin{array}{l} 2 \\ 1 \end{array}\right) I_1\left[\psi^{(1)}(1)-\psi^{(1)}\left(\frac{1}{2}\right)\right] \\ &+\left(\begin{array}{l} 2 \\ 2 \end{array}\right) I_2\left[\psi(1)-\psi\left(\frac{1}{2}\right)\right] \\ =&\frac{\pi}{2}(12)\zeta(3)-2 \pi \ln 2\left(-\frac{\pi^{2}}{3}\right)+\left(\frac{\pi^{3}}{6}+2 \pi \ln ^{2}2\right) \ln 4 \\ =& 6 \pi \zeta(3)+\pi^{3} \ln 2+4 \pi \ln ^{3} 2 \end{aligned} $$

Though we can find the integral $I_n=\int_{0}^{\infty} \frac{\ln ^{n}\left(1+x^{2}\right)}{1+x^{2}}$ one by one by the reduction formula, it is hard to find $$ \boxed{\psi^{(n)}(1)-\psi^{(n)}\left(\frac{1}{2}\right)\textrm{ where }n\geq 2}, $$ which is guessed to be $r\zeta(n+1)$ for some real number $r$? Your help will be highly appreciated.

Latest Edit

Being reminded by KStarGamer that $$ \psi^{(n)}(1)-\psi^{(n)}\left(\frac{1}{2}\right)=(-1)^{n} n !\left(2^{n+1}-2\right)\zeta(n+1) \tag*{(**)} $$

I can now finally conclude that

$$\boxed{I_n= 2 \ln 2 I_{n-1}+ (n-1)!\sum_{k=0}^{n-2} \frac{2^{n-k}-2}{k!}\zeta(n-k) I_k} $$

where $n\geq 1.$

For completeness, I tried to prove the identity (**) by the series representation of Polygamma Function:

$$ \psi^{(n)}(z)=(-1)^{n+1} n ! \sum_{k=0}^{\infty} \frac{1}{(z+k)^{n+1}} $$$$ \begin{aligned} \psi^{(n)}(1) &=(-1)^{n+1} n ! \sum_{k=0}^{\infty} \frac{1}{(1+k)^{n+1}} \\ &=(-1)^{n+1} n ! \zeta(n+1) \end{aligned} $$ $$ \begin{aligned} \psi^{(n)}\left(\frac{1}{2}\right) &=(-1)^{n+1} n ! \sum_{k=0}^{\infty} \frac{1}{\left(\frac{1}{2}+k\right)^{n+1}} \\ &=(-1)^{n+1} n ! 2^{n+1} \sum_{k=0}^{\infty} \frac{1}{(2 k+1)^{n+1}}\\&= (-1)^{n+1} n !\left(2^{n+1}-1\right) \zeta(n+1) \end{aligned} $$

Hence $$ \psi^{(n)}(1)-\psi^{(n)}\left(\frac{1}{2}\right)=(-1)^{n} n !\left(2^{n+1}-2\right)\zeta(n+1) $$