Find the integral
$$\int_{0}^{\infty}\frac{\tan^2{x}}{x^2}\,dx=? \tag{1}$$
and
$$\int_{0}^{\infty}\frac{x^2}{\tan^2{x}}\,dx=?\tag{2}$$
I know this
$$\int_{0}^{\infty}\frac{\sin^2{x}}{x^2}\,dx=\frac{\pi}{2}$$
and
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{\sin^2{x}}\,dx=\pi\ln{2}$$
But for $(1)$ and $(2)$ I can't find it. Thank you someone can find it.
And the wolf can't work.
Neither of your integrals exist. The integrands are positive and have (non-integrable) singularities at points where $\tan^2 x$ is $+\infty$ and $0$ respectively.
here is the description by photo why both integral not converge on ${0\to\infty}$
here is why ?
for the $\int_{0}^{\infty}\frac{x^2 }{\tan^2x}dx$
its the area under the blue line from $0$ to $\infty$
if you look at the picture its seem that the area=$\infty$
for the $\int_{0}^{\infty}\frac{\tan^2 x}{x^2}dx$
also it seem that the area=$\infty$
\begin{align} \int_{0}^{\pi/2} {\tan^{2}\left(x\right) \over x^{2}}\,{\rm d}x &= \int_{-\pi/2}^{0} {\tan^{2}\left(x + \pi/2\right) \over \left(x + \pi/2\right)^{2}}\,{\rm d}x = \int^{\pi/2}_{0} {\cot^{2}\left(x\right) \over \left(x - \pi/2\right)^{2}}\,{\rm d}x \sim {1 \over x}\quad\mbox{when}\quad x \sim 0 \end{align}
So, what ?
