Integrate logarithmic function $\int_0^\infty \frac{\ln(1+25x^2)}{1+16x^2} dx$

Question

Integration of logarithmic function from $0$ to infinity $$\int_0^\infty \frac{\ln(1+25x^2)}{1+16x^2} dx$$

My approach:

I substituted $x=\frac{1}{4}\tan z$, doing so denominator term reduced, bit still an integral is there in numerator which is difficult to evaluate.

Answer 1

Let $\tan t=4x$. Then,

$$\int_0^\infty \frac{\ln(1+25x^2)}{1+16x^2} dx =\frac14\int_0^{\pi/2}\ln (1+\frac{25}{16}\tan^2 t) dt = \frac14 I\left(\frac{25}{16}\right)$$

where

$$ I(a)= \int_0^{\pi/2}\ln (1+a \tan^2t)dt,\>\>\> I’(a) = \int_0^{\pi/2}\frac{\tan^2t dt}{1+a\tan^2t}= \frac{\pi}{2\sqrt a(\sqrt a+1)} $$

Thus,

$$\int_0^\infty \ln \frac{1+25x^2}{1+16x^2} dx = \frac14 I\left(\frac{25}{16}\right)=\frac14\int_0^{\frac{25}{16}}I’(a)da\\ =\frac\pi8\int_0^{25/16} \frac{da}{\sqrt a(\sqrt a+1)} =\frac\pi4\ln(\sqrt a+1)\bigg|_0^{\frac{25}{16} } = \frac\pi4\ln \frac94$$

Answer 2

Let $ a,b $ be reals such that $ 0<b<a $, we have the following :

\begin{aligned}\int_{0}^{+\infty}{\frac{\ln{\left(1+a^{2}x^{2}\right)}}{1+b^{2}x^{2}}\,\mathrm{d}x}&=\int_{0}^{+\infty}{\int_{0}^{1}{\frac{a^{2}x^{2}}{\left(1+b^{2}x^{2}\right)\left(1+a^{2}x^{2}y\right)}\mathrm{d}y}\,\mathrm{d}x}\\&=\int_{0}^{+\infty}{\int_{0}^{1}{\left(\frac{1}{\left(y-\frac{b^{2}}{a^{2}}\right)\left(1+b^{2}x^{2}\right)}-\frac{1}{\left(y-\frac{b^{2}}{a^{2}}\right)\left(1+a^{2}x^{2}y\right)}\right)\mathrm{d}y}\,\mathrm{d}x}\\ &=\int_{0}^{+\infty}{\int_{0}^{1}{\frac{\mathrm{d}y\,\mathrm{d}x}{\left(y-\frac{b^{2}}{a^{2}}\right)\left(1+b^{2}x^{2}\right)}}}-\int_{0}^{+\infty}{\int_{0}^{1}{\frac{\mathrm{d}y\,\mathrm{d}x}{\left(y-\frac{b^{2}}{a^{2}}\right)\left(1+a^{2}x^{2}y\right)}}}\\ &=\left(\int_{0}^{+\infty}{\frac{\mathrm{d}x}{1+b^{2}x^{2}}}\right)\left(\int_{0}^{1}{\frac{\mathrm{d}y}{y-\frac{b^{2}}{a^{2}}}}\right)-\int_{0}^{1}{\frac{1}{y-\frac{b^{2}}{a^{2}}}\int_{0}^{+\infty}{\frac{\mathrm{d}x}{1+a^{2}x^{2}y}}\,\mathrm{d}y}\\ &=\frac{\pi}{2b}\ln{\left(\frac{a^{2}}{b^{2}}-1\right)}-\frac{\pi}{a}\int_{0}^{1}{\frac{\mathrm{d}y}{2\sqrt{y}\left(y-\frac{b^{2}}{a^{2}}\right)}}\\ &=\frac{\pi}{2b}\ln{\left(\frac{a^{2}}{b^{2}}-1\right)}-\frac{\pi}{a}\int_{0}^{1}{\frac{\mathrm{d}y}{x^{2}-\frac{b^{2}}{a^{2}}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \textrm{We substituted : }y=x^{2}\\ &=\frac{\pi}{2b}\ln{\left(\frac{a^{2}}{b^{2}}-1\right)}-\frac{\pi}{2b}\left(\int_{0}^{1}{\frac{\mathrm{d}y}{x-\frac{b}{a}}}-\int_{0}^{1}{\frac{\mathrm{d}y}{x+\frac{b}{a}}}\right)\\ &=\frac{\pi}{2b}\ln{\left(\frac{a^{2}}{b^{2}}-1\right)}-\frac{\pi}{2b}\ln{\left(\frac{\frac{a}{b}-1}{\frac{a}{b}+1}\right)}\\ \int_{0}^{+\infty}{\frac{\ln{\left(1+a^{2}x^{2}\right)}}{1+b^{2}x^{2}}\,\mathrm{d}x}&=\frac{\pi}{b}\ln{\left(1+\frac{a}{b}\right)}\end{aligned}