Evaluate $\int_0^{\infty}\frac{\ln(x+\frac{1}{x})}{1+x^2}\cdot dx$

Question

Evaluate: $$\int_0^{\infty}\dfrac{\ln(x+\frac{1}{x})}{1+x^2}\cdot dx$$

I tried substituting $\ln(x+\frac{1}{x})=t$ that to some extent created denominator but made it more tedious. Please give some hints to solve.

Answer 1

Let $\displaystyle I = \int^{\infty}_{0}\ln(x+x^{-1})\cdot \frac{1}{1+x^2}dx$

Put $x=\tan \theta$ and $dx = \sec^2 \theta $ and changing limits

So $\displaystyle I = \int^{\frac{\pi}{2}}_{0}\frac{\ln(\tan \theta+\cot \theta)}{1+\tan^2 \theta}\cdot \sec^2 \theta d \theta = \int^{\frac{\pi}{2}}_{0}\ln(\tan \theta+\cot \theta)d\theta$

$\displaystyle I = \int^{\frac{\pi}{2}}_{0}\ln(2)-\int^{\frac{\pi}{2}}_{0}\ln (\sin2\theta)d\theta = \frac{\pi}{2}\ln(2)-\int^{\pi}_{0}\ln(\sin \theta)d\theta$

Above $\displaystyle \int^{\pi}_{0}\ln(\sin x)dx = \int^{\frac{\pi}{2}}_{0}\ln(\sin x)dx = -\pi \ln(2)$ is well known Integral

So $\displaystyle I = \frac{\pi}{2}\ln(2)+\frac{\pi}{2}\ln(2) = \pi\ln(2)$

Evaluation of $\displaystyle \int^{\frac{\pi}{2}}_{0}\ln(\sin x)dx$

$$I=\int_{0}^{\frac{\pi}{2}} \ln (\sin u) du$$

And the substitution $v=\frac{\pi}{2}-x$ reduces the integral to,

$$I=\int_{0}^{\frac{\pi}{2}} \ln (\cos v) dv$$

$$I=\int_{0}^{\frac{\pi}{2}} \ln (\cos u) du$$

Now adding the integrals and noting properties of logarithms we have,

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

Double angle,

$$2I=\int_{0}^{\frac{\pi}{2}} \ln (\sin 2x) dx -\frac{\pi}{2} \ln 2$$

The substitution $s=2x$ gives $$2I=\frac{1}{2}\int_{0}^{\pi} \ln (\sin s) ds -\frac{\pi}{2} \ln 2$$

But

$$\int_{0}^{\pi} \ln (\sin s) ds=2I$$

Follows from the substitution $w=\frac{\pi}{2}-s$ and the evenness of the function $f(w)=\ln (\cos w)$:

$$\int_{0}^{\pi} \ln (\sin s) ds$$ $$=-\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \ln (\cos w) dw$$

$$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln (\cos w)dw $$ $$=2 \int_{0}^{\frac{\pi}{2}} \ln (\cos w) dw=2I$$

So,

$$2I=I-\frac{\pi}{2}\ln 2$$

$$I=-\frac{\pi}{2}\ln 2$$