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Here, I am calculating $\int_{0}^{\frac{\pi}{2}} \frac{\tan x}{x}\,dx$ by parts: $\int_a^b u\,dv=uv|_a^b-\int_a^bv~~du$, in which $ dv=\frac{1}{x},~v=\ln x ,~u=\tan x,~du=\sec^2 x$.

I can derive that $$\tan x\ln x|_0^{\frac{\pi}{2}}-\int_0^\frac{\pi}{2}\ln x \sec^2 x\,dx.$$

But I can see that $\tan x$ is not defined at $x=\frac{\pi}{2}$ and $\ln x$ is not defined at $x=0$.

I cannot go further from here. So, I want to evaluate this integral problem. Any help is appreciated.

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  • $\begingroup$ This integral does not converge. $\endgroup$ – Euler....IS_ALIVE Oct 12 '18 at 11:33
  • $\begingroup$ Yes, I realized that. Thanks @Euler....IS_ALIVE $\endgroup$ – kunarapu priyatham Oct 12 '18 at 12:13
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You have this inequality:

$$\int_0^{\pi/2} \frac{\tan x}{x} \; dx >\int_1^{\pi/2} \frac{\tan x}{x} \; dx > \int_1^{\pi/2} \frac{\tan x}{1} \; dx.$$

This last integral is easily shown to be infinity.

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$$I=\int_0^{\pi/2}\frac{\tan(x)}{x}dx$$ we know that: $$\tan(x)=\sum_{n=0}^\infty\frac{(-1)^n2^{2n+2}\left(2^{2n+2}-1\right)B_{2n+2}}{(2n+2)!}x^{2n+1}$$ so we can say that: $$\int\frac{\tan(x)}{x}=\int\sum_{n=0}^\infty\frac{(-1)^n2^{2n+2}\left(2^{2n+2}-1\right)B_{2n+2}}{(2n+2)!}x^{2n}dx$$ which can easily be solved. However, if you look at the $\tan(\pi/2)$ it diverges to $\infty$n as: $$\lim_{x\to(\pi/2)^+}\tan(x)=\infty$$ so for this value the integral is also not defined, and is divergent.

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