I recently attempted to evaluate the following integral
$$\int_{-\infty}^\infty\frac{\ln{(x^4+x^2+1)}}{x^4+1}dx$$
I started by inserting a parameter, $t$
$$F(t)=\int_{-\infty}^\infty\frac{\ln{(tx^4+x^2+t)}}{x^4+1}dx$$
Where F(0) is the following
$$F(0)=\int_{-\infty}^\infty\frac{\ln{(x^2)}}{x^4+1}dx=2\int_0^\infty\frac{\ln{(x^2)}}{x^4+1}dx=4\int_0^\infty\frac{\ln x}{x^4+1}dx$$
We can evaluate this using a common integral from complex analysis and taking the derivative using Leibniz’s rule.
$$\int_0^\infty\frac{x^m}{x^n+1}dx=\frac{1}{m+1}\int_0^\infty\frac{(m+1)x^m}{(x^{m+1})^\frac{n}{m+1}+1}dx=\frac{1}{m+1}\int_0^\infty\frac{du}{x^\frac{n}{m+1}+1}$$
$$=\frac{1}{m+1}\frac{\pi}{\frac{n}{m+1}\sin{\frac{\pi}{\frac{n}{m+1}}}}=\frac{\pi}{n\sin{\frac{\pi(m+1)}{n}}}=\frac{\pi}{n}\csc{\frac{\pi(m+1)}{n}}$$
$$\int_0^\infty\frac{\ln{x}}{x^n+1}dx=\frac{d}{dm}\int_0^\infty\frac{x^m}{x^n+1}dx\Big|_{m=0}$$
$$=\frac{\pi}{n}\frac{d}{dm}\csc{\frac{\pi(m+1)}{n}}\Big|_{m=0}=-\frac{\pi^2}{n^2}\csc{\frac{\pi(m+1)}{n}}\cot{\frac{\pi(m+1)}{n}}\big|_{m=0}=-\frac{\pi^2}{n^2}\csc{\frac{\pi}{n}}\cot{\frac{\pi}{n}}$$
Therefore
$$F(0)=4\int_0^\infty\frac{\ln{x}}{x^4+1}dx=-\frac{\pi^2\sqrt 2}{4}=-\frac{\pi^2}{2\sqrt 2}$$
Now that we found F(0), we can start applying Feynman’s trick.
$$F’(t)=\int_{-\infty}^{\infty}\frac{dx}{tx^4+x^2+t}$$
Using a formula I derived we can continue
$$\int_{-\infty}^\infty\frac{dx}{ax^4+bx^2+c}=\frac{\pi}{\sqrt{c}\sqrt{b+2\sqrt{ac}}}$$
$$F’(t)=\frac{\pi}{\sqrt{t}\sqrt{1+2t}}$$
Integrating both sides
$$F(t)=\pi\sqrt2\ln{(\sqrt{2t}+\sqrt{2t+1})}+C$$
Set $t=0$
$$C=F(0)=-\frac{\pi^2}{2\sqrt2}$$
Therefore
$$F(t)=\pi\sqrt2\ln{(\sqrt{2t}+\sqrt{2t+1})}-\frac{\pi^2}{2\sqrt2}$$
$$I=\pi\sqrt2\ln{(\sqrt2+\sqrt3)}-\frac{\pi^2}{2\sqrt2}$$
WolframAlpha confirms it numerically
I am not satisfied with this solution. I am curious as to what other solutions there might be. How else can we solve this integral?
Best Answer
Utilize the integral $\int_{-\infty}^\infty \frac{\ln(x^2+a^2)}{x^2+b^2}dx=\frac{2\pi}b \ln(a+b)$, along with the shorthands $p=e^{i\frac\pi6}$ and $q= e^{-i\frac\pi4} $ \begin{align} &\int_{-\infty}^\infty\frac{\ln{(x^4+x^2+1)}}{x^4+1}dx\\ =& \int_{-\infty}^\infty\frac{\ln{(x^2+p^2) (x^2+\bar{p}^2)}}{(x^2+q^2)( x^2+\bar{q}^2) }dx\\ =& \ \Im \int_{-\infty}^\infty \frac{\ln(x^2+p^2)}{x^2+{q}^2} + \frac{\ln(x^2+\bar{p}^2)}{x^2+{q}^2} \ dx\\ =& \ \Im \frac{2\pi}{q}\left[\ln (p+q)+\ln(\bar p+q)\right]\\ = &\ \sqrt2\pi\left( \ln(\sqrt2+\sqrt3)-\frac\pi4\right) \end{align}