Integral $\int_{0}^{1}\frac{\ln\left(x^2\right)}{\left(1+x^{2}\right)\left(\pi^{2}+\ln^{2}x\right)}dx$

Question

$$I=\int_{0}^{1}\frac{\ln\left(x^2\right)}{\left(1+x^{2}\right)\left(\pi^{2}+\ln^{2}x\right)}dx$$ $$I=2\int_{0}^{1}\frac{\ln\left(x\right)}{\left(1+x^{2}\right)\left(\pi^{2}+\ln^{2}x\right)}dx$$ Substituting $x\to\frac{1}{x}$: $$I=-2\int_{1}^{\infty}\frac{\ln\left(x\right)}{\left(x^{2}+1\right)\left(\pi^{2}+\ln^{2}x\right)}dx$$ This would imply: $$\int_{0}^{\infty}\frac{\ln\left(x\right)}{\left(1+x^{2}\right)\left(\pi^{2}+\ln^{2}x\right)}dx=0$$

After searching the Integral on Approach0, I found various sources where an Incorrect Value is given: $$\int_{0}^{1}\frac{\ln\left(x^2\right)}{\left(1+x^{2}\right)\left(\pi^{2}+\ln^{2}x\right)}dx=\ln2-\frac{1}{2}$$ This is wrong, as can be checked numerically. But a user on AoPS gave the following Correction (AoPS) with the Solution as Well: $$\int_{0}^{1}\frac{\ln\left(x^2\right)}{\left(1\color{red}{-}x^{2}\right)\left(\pi^{2}+\ln^{2}x\right)}dx=\frac{1}{2}-\ln2$$

Trying out the Solution Step by Step, I was able to get till here: $$I=-2\int_{0}^{\infty}e^{-\pi u}\sum_{n=0}^{\infty}\left(\left(-1\right)^{n}\frac{u}{u^{2}+\left(2n+1\right)^{2}}\right)du$$

Now the problem is that factor of $(-1)^n$ which is not present in the corrected problem, without it the sum evaluates to: $$\sum_{n=0}^{\infty}\left(\frac{u}{u^{2}+\left(2n+1\right)^{2}}\right)=\frac{\pi}{4}\tanh\left(\frac{\pi u}{2}\right)$$ But I am not sure how to evaluate it with the $(-1)^n$ factor.
Wolfram only gives a Partial Sum Formula in terms of Lerch Transcendent.

In the Post Here (AoPS), the Integral has a Series form as follows: $$I=-\frac{4}{\pi}\sum_{m=0}^{\infty}\left(\frac{\left(-1\right)^{m}\left(2m+1\right)}{\left(2m+1\right)^{2}-4}\ln\left(m+\frac{1}{2}\right)\right)$$

Another Post where the Wrong Integral Equality is given: Problem #67
I also think I saw this Integral before on MSE, but I am not able to find it.

Answer 1

Let,
$$\qquad \text{Si}(z):=\int_0^{z}\frac{\sin(t)}{t}\mathrm{d}t\qquad\text{Ci}(z)=\gamma+\ln(z)+\int_0^{z}\frac{\cos(t)-1}{t}\mathrm{d}t$$ the sine and cosine integral functions.
Where $\gamma=-\Gamma'(1)$ is the Euler-Mascheroni constant

$$\int_{0}^{1}\frac{\ln\left(x^{2}\right)}{\left(1+x^{2}\right)\left(\pi^{2}+\ln\left(x\right)^{2}\right)}dx$$ $$2\int_{0}^{1}\frac{\ln\left(x\right)}{\left(1+x^{2}\right)\left(\pi^{2}+\ln\left(x\right)^{2}\right)}dx$$ $$\ln(x)=t, x=e^t, dx=e^t dt$$ $$2\int_{-\infty}^{0}\frac{te^{t}}{\left(1+e^{2t}\right)\left(\pi^{2}+t^{2}\right)}dt$$ $$-\int_{0}^{\infty}\frac{t}{\pi^{2}+t^{2}}\operatorname{sech}\left(t\right)dt$$ $$-\int_{0}^{\infty}\frac{t}{\pi^{2}+t^{2}}2e^{-t}\sum_{k=0}^{\infty}\left(-1\right)^{k}e^{-2kt}dt$$ $$-2\sum_{k=0}^{\infty}\left(-1\right)^{k}\int_{0}^{\infty}\frac{t}{\pi^{2}+t^{2}}e^{-\left(2k+1\right)t}dt$$ For this integral I used Wolfram. $$-2\sum_{k=0}^{\infty}\left(-1\right)^{k}\left(\text{Ci}\left(\left(2k+1\right)\pi\right)\left(-\cos\left(\left(2k+1\right)\pi\right)\right)+\frac{\left(\pi-2\text{Si}\left(\left(2k+1\right)\pi\right)\right)\sin\left(\left(2k+1\right)\pi\right)}{2}\right)$$ But $$\cos((2k+1)\pi)=-1\text{ and }\sin((2k+1)\pi)=0$$ So you have $$\int_{0}^{1}\frac{\ln\left(x^{2}\right)}{\left(1+x^{2}\right)\left(\pi^{2}+\ln\left(x\right)^{2}\right)}dx=-2\sum_{k=0}^{\infty}\left(-1\right)^{k}\text{Ci}\left(\left(2k+1\right)\pi\right)$$

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Edit
An other method can be using Feymann trick:
You have to calculate this: $$-\int_{0}^{\infty}\frac{t}{\pi^{2}+t^{2}}\operatorname{sech}\left(t\right)dt$$ You can consider $$f(x)=\int_{0}^{\infty}\frac{t}{x^{2}+t^{2}}\operatorname{sech}\left(t\right)dt$$ So you have that the integral of $f(x)$ is $$F(x)=\int_{0}^{\infty}\arctan\left(\frac{x}{t}\right)\operatorname{sech}\left(t\right)dt$$ You can try to evaluate $F(x)$ and then you can find $F'(\pi)$ (your constant)

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Edit 1:
Thanks to @Zima,
You can express it as
$$\ln(2)-\frac{1}{2}-4\sum_{k=0}^{\infty}\text{Ci}((4k+1)\pi)$$ Where $$-4\sum_{k=0}^{\infty}\text{Ci}((4k+1)\pi)=\int_0^{\infty}e^{-\pi x}\Im\left[\psi\left(\frac{1+ix}{4}\right)\right]\mathrm{d}x$$

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Edit 2:
Some similar series: $$\sum_{n=1}^{\infty}\text{Ci}(n\pi)=\frac{\ln(2)-\gamma}{2}\qquad \sum_{n=1}^{\infty}(-1)^n\text{Ci}(2n\pi)=1-\ln(2)-\frac{\gamma}{2}$$

Answer 2

Composite method to approximate the function :

The function in the integral is hard with some polynomials,rational function,or nth root function to approximate here is my try :

$$h(x)=2 (-(1-x^{1/3}) (1-(1/(x^{1/3}+1)))+10/18 (1-x^{1/2}) (1-1/(x+1))))-1/12 (1-x)^2 \sqrt{x}+1/12 (1-x)^4 x-1/12 (1-x)^5 x^2-1/12 (1-x)^2 x-1/12 (1-x)^2 x^2,f(x)=\frac{\ln\left(x^2\right)}{\left(1+x^{2}\right)\left(\pi^{2}+\ln^{2}x\right)},r(x)=-15 (x (2-x^2-(1/\sqrt{x}))) (1-\sqrt{x}))^2,s(x)=(1-\sqrt{x}) (r(x)-r((55/1000))),h_1(x)=13 (1-x)^2 (3/5-x)^2 x^2$$

Then :

$$10 (h(x)-f(x))\simeq s(x)+h_1(x)$$

Hoping someone can pursue it .

Another idea to tackle it :

Let :

$$f(x)=\sqrt{2} (x^{1/3} (1-x^{1/5}) (1-x)^{1/4}+1/12 x^{1/5} (1-x)^2+1/18 x^{1/7} (1-x)^4+ 1/9x^{1/9} (1-x)^6+1/25 x^{1/11} (1-x)^8+\cdots)$$

$$g(x)=-\sqrt{3}\frac{\ln\left(x^2\right)}{\left(1+x^{2}\right)\left(\pi^{2}+\ln^{2}x\right)}$$

Then the idea is to delete one extrema making the difference and divide the difference in sector (two stationary one linear ) and use power series at judicious point which can be found with a graph .