Evaluate : $I=\int_0^{\infty}\frac{\ln (1+ax+x^{2})}{1+x^2}\,dx$

Question

Does the following integral have a closed form :

$$I=\displaystyle\int\limits_0^{\infty}\frac{\ln (1+ax+x^{2})}{1+x^2}dx$$

Where $|a|≤1$ ,

I was trying using Feynman's trick.

Define

$$I(b)=\displaystyle\int\limits_0^{\infty}\frac{\ln (b(1+x^{2})+ax)}{1+x^2}dx$$

Differentiating with respect to $b$ we get :

$$I'(b)=\displaystyle\int\limits_0^{\infty}\frac{1}{b+ax+bx^{2}}dx$$

$$=2\left(\frac{π}{2\sqrt{4b^{2}-a^{2}}}-\frac{\arctan \left(\frac{a}{\sqrt{4b^{2}-a^{2}}}\right)}{\sqrt{4b^{2}-a^{2}}}\right)$$

Known :

$$\displaystyle\int \frac{1}{\sqrt{4b^{2}-a^{2}}}db=\frac{\log (2 x + \sqrt{-a^2 + 4 x^{2}})}{2}$$

my problem in this integral :

$$\displaystyle\int\limits_0^{1}\frac{\arctan \left(\frac{a}{\sqrt{4b^{2}-a^{2}}}\right)}{\sqrt{4b^{2}-a^{2}}}db=?$$

Of course here

$$I=I(1)=I(0)+\int\limits_0^{1}I'(b)db$$

$$I(0)=\frac{π\ln a}{2}$$

I already waiting your hints or solution.

Answer 1

The trick to solving this integral is to reduce it to two power series and then evaluate the power series using differential equations. The final answer is

$$\int_{0}^{\infty}\frac{\ln\left(1+ax+x^{2}\right)}{1+x^{2}}dx=\left(\frac{1}{2}\sin^{-1}\left(\frac{a}{2}\right)-\frac{\pi}{4}\right)\ln\left(\frac{1-\sqrt{1-\frac{a^{2}}{4}}}{1+\sqrt{1-\frac{a^{2}}{4}}}\right)+\frac{\pi}{2}\ln\left|a\right|-\Delta_{\pi}\mathrm{Cl}_{2}\left(\sin^{-1}\left(\frac{a}{2}\right)\right)$$

Here, $\Delta_h$ is the forward difference operator defined by

$$\Delta_h[f](x)=f(x+h)-f(x)$$

And $\mathrm{Cl}_2(\theta)$ is the SL-type Clausen function, defined by

$$\mathrm{Cl}_2(\varphi)=\int_0^\varphi \ln\left|2\sin\left(\frac{x}{2}\right)\right|dx$$

If you want to compute answers to your integral, you can use the fact that

$$\Delta_{\pi}\mathrm{Cl}_{2}\left(\sin^{-1}\left(\frac{a}{2}\right)\right)=-2\sum_{k=0}^{\infty}\frac{\sin\left(\left(2k+1\right)\sin^{-1}\left(\frac{a}{2}\right)\right)}{\left(2k+1\right)^{2}}$$

To begin solving the integral, we substitute $x=\tan(\theta)$ to get that

\begin{align*} I&=\int_{0}^{\infty}\frac{\ln\left(1+ax+x^{2}\right)}{1+x^{2}}dx\\ &=\int_{0}^{\frac{\pi}{2}}\ln\left(\sec^{2}\left(x\right)+a\tan\left(x\right)\right)dx\\ &=\int_{0}^{\frac{\pi}{2}}\ln\left(1+a\sin\left(x\right)\cos\left(x\right)\right)dx-2\int_{0}^{\frac{\pi}{2}}\ln\left(\cos\left(x\right)\right)dx\\ \end{align*}

We can now use the fact that $\sin(2x)=2\sin(x)\cos(x)$ and symmetries in the argument of $\sin(x)$ to get that

$$\int_{0}^{\frac{\pi}{2}}\ln\left(1+a\sin\left(x\right)\cos\left(x\right)\right)dx=\int_{0}^{\frac{\pi}{2}}\ln\left(1+\frac{a}{2}\sin\left(x\right)\right)dx$$

We will also note that it is very easy to show that

$$2\int_{0}^{\frac{\pi}{2}}\ln\left(\cos\left(x\right)\right)dx=-\pi\ln\left(2\right)$$

We can now substitute these two expressions into where we left off, preparing to expand $\ln(x)$ as a power series, to get that

\begin{align*} I&=\int_{0}^{\frac{\pi}{2}}\ln\left(1+\frac{a}{2}\sin\left(x\right)\right)dx+\pi\ln\left(2\right)\\ &=-\int_{0}^{\frac{\pi}{2}}\sum_{n=1}^{\infty}\frac{\left(-\frac{a}{2}\sin\left(x\right)\right)^{n}}{n}dx+\pi\ln\left(2\right)\\ &=-\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n}\left(\frac{a}{2}\right)^{n}\int_{0}^{\frac{\pi}{2}}\left(\sin\left(x\right)\right)^{n}dx+\pi\ln\left(2\right) \end{align*}

We can now use the well known result that

$$\int_{0}^{\frac{\pi}{2}}\left(\sin\left(x\right)\right)^{n}dx=\frac{\sqrt{\pi}\Gamma\left(\frac{n+1}{2}\right)}{2\Gamma\left(\frac{n}{2}+1\right)}$$

where $\Gamma(x)=(x-1)!$ is the gamma function to get that

\begin{align*} I&=-\frac{\sqrt{\pi}}{2}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n}\left(\frac{a}{2}\right)^{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}+\pi\ln\left(2\right)\\ &=-\frac{\sqrt{\pi}}{4}\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{a}{2}\right)^{2n}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n+1\right)}+\frac{\sqrt{\pi}}{2}\sum_{n=1}^{\infty}\frac{1}{2n+1}\left(\frac{a}{2}\right)^{2n+1}\frac{\Gamma\left(n+1\right)}{\Gamma\left(n+1+\frac{1}{2}\right)}+\pi\ln\left(2\right) \end{align*}

Where the last equality was obtained by summing over even/odd numbers. We can now use the Legendre duplication formula, namely

$$\Gamma\left(n+\frac{1}{2}\right)=\sqrt{\pi}2^{1-2n}\frac{\Gamma\left(2n\right)}{\Gamma\left(n\right)}$$

to get both of the series in terms of factorials, namely

$$I=-\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{1}{n}\frac{\left(2n\right)!}{\left(n!\right)^{2}}\left(\frac{a}{4}\right)^{2n}+\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}\frac{\left(n!\right)^{2}}{\left(2n\right)!}a^{2n+1}+\pi\ln\left(2\right)$$

The first power series we turn our attention to is

$$\sum_{n=1}^{\infty}\frac{\left(2n\right)!}{\left(n!\right)^{2}}x^{n}$$

We can use the fact that

$$\frac{\left(2(n+1)\right)!}{\left((n+1)!\right)^{2}}=\frac{\left(2n\right)!}{\left(n!\right)^{2}}\cdot\left(\frac{4n+2}{n+1}\right)$$

to set up an easily solvable differential equation, which yields

$$\sum_{n=1}^{\infty}\frac{\left(2n\right)!}{\left(n!\right)^{2}}x^{n}=\frac{1}{\sqrt{1-4x}}-1$$

And thus by integrating

$$\sum_{n=1}^{\infty}\frac{1}{n}\frac{\left(2n\right)!}{\left(n!\right)^{2}}x^{n}=\ln\left(\frac{1-\sqrt{1-4x}}{1+\sqrt{1-4x}}\right)-\ln\left(x\right)$$

Substituting back in yields

$$I=-\frac{\pi}{4}\ln\left(\frac{1-\sqrt{1-\frac{a^{2}}{4}}}{1+\sqrt{1-\frac{a^{2}}{4}}}\right)+\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}\frac{\left(n!\right)^{2}}{\left(2n\right)!}a^{2n+1}+\frac{\pi}{2}\ln\left(a\right)$$

Similarly, it can be shown that

$$\sum_{n=0}^{\infty}\frac{\left(n!\right)^{2}}{\left(2n\right)!}x^{2n+1}=\frac{4\left(x\sqrt{4-x^{2}}+\sin^{-1}\left(\frac{x}{2}\right)x^{2}\right)}{\left(4-x^{2}\right)^{\frac{3}{2}}}$$

Which upon integrating once yields

$$\sum_{n=0}^{\infty}\frac{1}{2n+1}\frac{\left(n!\right)^{2}}{\left(2n\right)!}x^{2n+1}=\frac{4\sin^{-1}\left(\frac{x}{2}\right)}{\sqrt{4-x^{2}}}$$

And then integrating once more and substituting into our equation yields the final solution. If you have any questions about any of the steps, since I skipped over a lot, you can ask me.

Answer 2

Parameterising $2\sin a$ instead of $a$ yields the integral $$ I(a)=\int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x, $$ where $a\in (0, \frac{\pi}{2}). $

Differentiating $I(a)$ w.r.t. $a$ yields $$ \begin{aligned} I^{\prime}(a) &=\int_{0}^{\infty} \frac{2 x \cos a}{\left(1+x^{2}\right)\left(1+2 x \sin a+x^{2}\right)} d x \\ &=\cot a\int_{0}^{\infty}\left(\frac{1}{1+x^{2}}-\frac{1}{1+2 x \sin a+x^{2}}\right) d x \\ &=\cot a\left[\tan ^{-1} x-\frac{1}{\cos a} \tan ^{-1}\left(\frac{x+\sin a}{\cos a}\right)\right]_{0}^{\infty} \\ &=\cot a\left[\frac{\pi}{2}-\frac{1}{\cos a}\left(\frac{\pi}{2}-a\right)\right] \end{aligned} $$

Integrating $I’(a)$ back to $I(a)$, we have

$$ \begin{aligned} I\left(a\right)- \underbrace{I(0)}_{=\pi\ln 2} &=\int_{0}^{a} \cot x\left[\frac{\pi}{2}-\frac{1}{\cos x}\left(\frac{\pi}{2}-x\right)\right] d x \\ &=\frac{\pi}{2} \underbrace{ \int_{0}^{a}\left(\cot x-\frac{1}{\sin x}\right) d x}_{= 2 \ln \left(\cos \frac{a}{2}\right)} + \underbrace{\int_{0}^{a} \frac{x}{\sin x} d x}_{K} \end{aligned} $$

$$ \begin{aligned} K &=\int_{0}^{a} \frac{x}{\sin x} d x=\int_{0}^{a} x\, d\left[\ln \left(\tan \frac{x}{2}\right)\right] \\ &=\left[x\ln \left(\tan \frac{x}{2}\right)\right]_{0}^{a}-\int_{0}^{a} \ln \left(\tan \frac{x}{2}\right) d x \\ &=a \ln \left(\tan \frac{a}{2}\right)-2 \int_{0}^{\frac{a}{2}} \ln (\tan x) d x \end{aligned} $$

Now we can conclude that, for any $a\in (0, \frac{\pi}{2}]$, $$ \boxed{I=I\left(a\right)=\pi \ln 2+\pi \ln \left(\cos \frac{a}{2}\right)+a \ln \left(\tan \frac{a}{2}\right) -2 \int_{0}^{\frac{a}{2}} \ln (\tan x) d x} $$

Similarly, for any $a\in (-\frac{\pi}{2},0)$, $$ \boxed{I=I\left(a\right)=\pi \ln 2+\pi \ln \left(\cos \frac{a}{2}\right)-a \ln \left(\tan \frac{-a}{2}\right) +2 \int_{0}^{\frac{-a}{2}} \ln (\tan x) d x} $$

For the last integral, we may evaluate it by expressing it as a series of sine as below:

$$ \int_{0}^{x} \ln (\tan \theta) d \theta=-\sum_{k=0}^{\infty} \frac{\sin ((4 k+2) x)}{(2 k+1)^{2}} $$

from the web.

Example 1 $$\begin{aligned}\quad \int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x&= I\left(-\frac{\pi}{6}\right)\\&= \frac{\pi}{2}[\ln (2+\sqrt{3})]+\frac{\pi}{6} \ln (2+\sqrt{3})+2\left(-\frac{2}{3} G\right)\\& =\frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G \end{aligned}$$

Example 2

$$ \begin{aligned} &\int_{0}^{\infty} \frac{\ln \left(1+\sqrt{2} x+x^{2}\right)}{1+x^{2}}\\=&I\left(\frac{\pi}{4}\right) \\ =& \pi \ln \left(2 \cos \frac{\pi}{8}\right)-\frac{\pi}{4} \ln \left(\tan \frac{\pi}{8}\right)-2 \int_{0}^{\frac{\pi}{8}} \ln (\tan x) d x \\ =& \frac{\pi}{2} \ln (2+\sqrt{2})+\frac{\pi}{4} \ln (\sqrt{2}+1) -2\left[\frac{\pi}{8} \ln (\sqrt{2}-1)-\Im\left(\operatorname{Li}_{2}(i(\sqrt{2}-1))\right]\right.\\=& \pi \ln [\sqrt[4]{2}(\sqrt{2}+1)] +2 \Im\left(\operatorname{Li}_{2}(i(\sqrt{2}-1))\right. \end{aligned} $$ where the last integral see post

The answer is not satisfactory as the last integral is hard to evaluate!