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how to solve $$\mathcal{J(a)}=\int _0^1\frac{\ln \left(1+x\right)}{a^2+x^2}\:\mathrm{d}x$$

i used the differentiation under the integral and got

\begin{align} \mathcal{J(b)}&=\int _0^1\frac{\ln \left(1+bx\right)}{a^2+x^2}\:\mathrm{d}x \\[3mm] \mathcal{J'(b)}&=\int _0^1\frac{x}{\left(a^2+x^2\right)\left(1+bx\right)}\:\mathrm{d}x \\[3mm] &=\frac{a^2b}{1+a^2b^2}\int _0^1\frac{1}{a^2+x^2}\:\mathrm{d}x+\frac{1}{1+a^2b^2}\int _0^1\frac{x}{a^2+x^2}\:\mathrm{d}x-\frac{b}{1+a^2b^2}\int _0^1\frac{1}{1+bx}\:\mathrm{d}x \\[3mm] &=\frac{ab}{1+a^2b^2}\operatorname{atan} \left(\frac{1}{a}\right)+\frac{1}{2}\frac{\ln \left(1+a^2\right)}{1+a^2b^2}-\frac{\ln \left(a\right)}{1+a^2b^2}-\frac{\ln \left(1+b\right)}{1+a^2b^2} \end{align} But we know that $\mathcal{J}(1)=\mathcal{J(a)}$ and $\mathcal{J}(0)=0$ \begin{align} \int_0^1\mathcal{J'(b)}\:\mathrm{d}b&=a\:\operatorname{atan} \left(\frac{1}{a}\right)\int _0^1\frac{b}{1+a^2b^2}\:\mathrm{d}b+\frac{\ln \left(1+a^2\right)}{2}\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b-\ln \left(a\right)\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b \\ &-\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b \\[3mm] \mathcal{J(a)}&=\frac{1}{2a}\operatorname{atan} \left(\frac{1}{a}\right)\ln \left(1+a^2\right)+\frac{1}{2a}\ln \left(1+a^2\right)\operatorname{atan} \left(a\right)-\frac{1}{a}\ln \left(a\right)\:\operatorname{atan} \left(a\right)-\underbrace{\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b}_{\mathcal{I}} \end{align} but how to calculate ${\mathcal{I}}$, i tried using the same technique but it didnt work

  • Are you happy to use Complex Analysis or are you hoping to only use Real Methods? – David Galea Aug 13 '20 at 05:07
  • The best we can do is to express the integral in terms of Dilogarithms. First split the denominator using $\frac{1}{x^2 + a^2} = \frac{i}{2a}\left( \frac{1}{x+ia} - \frac{1}{x-ia}\right)$. Then, try to compute both integrals in terms of $\text{Li}_2(\cdot)$. It's a bit tedious but doable. – Shobhit Bhatnagar Aug 13 '20 at 05:17
  • @DavidGalea I'd prefer only real methods –  Aug 13 '20 at 05:23
  • At the end, you have $b$ used as the argument for $\mathcal{J}(b)$ and as the integration variable. Should it be integrated w.r.t. $x$ at the end? – Varun Vejalla Aug 13 '20 at 05:50
  • @VarunVejalla i dont think i understand what you are saying, i integrate with respect to x then with respect to b, thats just standard differentiation under the integral sign –  Aug 13 '20 at 06:21
  • You fixed that error now, but another notation issue is your double definition of $\mathcal{J}$. You should introduce a different notation for the second function to make it clearer. – Varun Vejalla Aug 13 '20 at 06:50

3 Answers3

1

Repeating your calculations using Feynman's trick $$J'(b)=\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)-2 \log (b+1)}{2 (a^2 b^2+1)}$$

$$J(b)=\int\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)}{2 (a^2 b^2+1)}\,db-\int\frac{\log(b+1)}{ a^2 b^2+1}\,db$$ The first integral is simple $$J_1(b)=\int\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)}{2 (a^2 b^2+1)}\,db$$ $$J_1(b)=\frac{\cot ^{-1}(a) \log \left(a^2 b^2+1\right)+\left(\log \left(a^2+1\right)-\log \left(a^2\right)\right) \tan ^{-1}(a b)}{2 a}$$ $$K_1=\int_0 ^1\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)}{2 (a^2 b^2+1)}\,db$$ $$K_1=\frac{\left(\log \left(a^2+1\right)-\log \left(a^2\right)\right) \tan ^{-1}(a)+\log \left(a^2+1\right) \cot ^{-1}(a)}{2 a}$$ More tedious is the second integral $$J_2(b)=\int\frac{\log(b+1)}{ a^2 b^2+1}\,db$$ $$J_2(b)=\frac{i \left(\text{Li}_2\left(\frac{a (b+1)}{a-i}\right)-\text{Li}_2\left(\frac{a (b+1)}{a+i}\right)+\log (b+1) \left(\log \left(1-\frac{a (b+1)}{a-i}\right)-\log \left(1-\frac{a (b+1)}{a+i}\right)\right)\right)}{2 a}$$ $$K_2(b)=\int_0^1\frac{\log(b+1)}{ a^2 b^2+1}\,db=$$ $$\frac{i \left(-\text{Li}_2\left(\frac{a}{a-i}\right)+\text{Li}_2\left(\frac{2 a}{a-i}\right)+\text{Li}_2\left(\frac{a}{a+i}\right)-\text{Li}_2\left(\frac{2 a}{a+i}\right)+\log (2) \left(\log \left(-\frac{a+i}{a-i}\right)-\log \left(-\frac{a-i}{a+i}\right)\right)\right)}{2 a}$$ which, for sure, is a real number.

Remark

I think that this could have benn done without Feynman's trick $$\frac{\log \left(1+x\right)}{a^2+x^2}=\frac{\log \left(1+x\right)}{(x+i a)(x-i a)}$$ $$\int \frac{\log \left(1+x\right)}{a^2+x^2}\,dx=\frac i {2a}\left(\int \frac{\log \left(1+x\right)}{x+i a}\,dx-\int \frac{\log \left(1+x\right)}{x-i a}\,dx \right)$$ $$\int \frac{\log \left(1+x\right)}{x+i a}\,dx=\text{Li}_2\left(\frac{x+1}{1-i a}\right)+\log (x+1) \log \left(1-\frac{x+1}{1-ia}\right)$$ $$\int \frac{\log \left(1+x\right)}{x-i a}\,dx=\text{Li}_2\left(\frac{x+1}{1+i a}\right)+\log (x+1) \log \left(1-\frac{x+1}{1+ia}\right)$$

Claude Leibovici
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0

The integral $I$ has been evaluated here by Felix Marin.

Using it yields the closed form of the integral \begin{align} J\left(a\right)&=\int_0^1\frac{\ln\left(1+x\right)}{a^2+x^2}\:dx=\frac{1}{2a}\operatorname{arctan} \left(\frac{1}{a}\right)\ln \left(1+a^2\right)+\frac{1}{2a}\ln \left(1+a^2\right)\operatorname{arctan} \left(a\right) \\ &-\frac{1}{a}\ln \left(a\right)\:\operatorname{arctan} \left(a\right)-2\ln \left(2\right)\frac{\operatorname{arctan} \left(a\right)}{a}-\frac{1}{a}\mathfrak{I}\left(\operatorname{Li}_2\left(\frac{2a}{i+a}\right)-\operatorname{Li}_2\left(\frac{a}{i+a}\right)\right) \end{align} Complex analysis seems like the only way to go with this integral because software also gives closed forms featuring the imaginary unit with polylogs, i also checked if this worked for particular values and it does, though i'd still like to see if anyone can come up with an approach without any complex methods.

Dennis Orton
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Suppose $\left(a,b,c,z\right)\in\mathbb{R}\times\mathbb{R}_{>0}\times\mathbb{R}_{\ge0}\times\mathbb{R}_{>0}$, and set

$$\alpha:=\arctan{\left(\frac{a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$

$$\gamma:=\arctan{\left(\frac{c-a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$

$$\theta:=\arctan{\left(\frac{z+a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

Then, the following logarithmic integral can be evaluated in terms of Clausen functions using the general result derived here:

$$\begin{align} \int_{0}^{z}\mathrm{d}x\,\frac{2b\ln{\left(x+c\right)}}{\left(x+a\right)^{2}+b^{2}} &=\operatorname{Cl}_{2}{\left(2\alpha+2\gamma\right)}-\operatorname{Cl}_{2}{\left(2\theta+2\gamma\right)}+\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi-2\theta\right)}\\ &~~~~~+\left(\theta-\alpha\right)\ln{\left(b^{2}\sec^{2}{\left(\gamma\right)}\right)},\\ \end{align}$$

where the Clausen function (of order 2) is defined for real arguments by the integral representation

$$\operatorname{Cl}_{2}{\left(\vartheta\right)}:=-\int_{0}^{\vartheta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\vartheta\in\mathbb{R}}.$$

Applying the formula to the particular case where $a=0\land c=1\land z=1$, and with a little help from the duplication formula for the Clausen function, we obtain the following delightfully compact result:

$$\begin{align} \forall b\in\mathbb{R}_{>0}:\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(x+1\right)}}{x^{2}+b^{2}} &=\frac{4\theta\ln{\left(\csc{\left(\theta\right)}\right)}-\operatorname{Cl}_{2}{\left(4\theta\right)}}{4b};~~~\small{\theta:=\arctan{\left(\frac{1}{b}\right)}\in\left(0,\frac{\pi}{2}\right)}.\\ \end{align}$$

It should be clear that Clausen functions are the natural tool to use if you're trying to stick to real methods.

David H
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