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(The Cauchy principal value of) $$ \int_0^{\infty}\frac{\tan x}{x}\mathrm dx $$

I tried to cut this integral into $$\sum_{k=0}^{\infty}\int_{k\pi}^{(k+1)\pi}\frac{\tan x}{x}\mathrm dx$$ And then $$\sum_{k=0}^{\infty}\lim_{\epsilon \to 0}\int_{k\pi}^{(k+1/2)\pi-\epsilon}\frac{\tan x}{x}\mathrm dx+\int_{(k+1/2)\pi+\epsilon}^{(k+1)\pi}\frac{\tan x}{x}\mathrm dx$$ $$\sum_{k=0}^{\infty}\int_{k\pi}^{(k+1/2)\pi}\frac{((2k+1)\pi-2x)\tan x}{((2k+1)\pi-x)x}\mathrm dx$$ And I did not know how to continue. I did not know if I was right or not. How to calculate this integral?

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You seem to be on the right track.

We have $$ P\int_0^\infty \frac{\tan{x}}{x}dx = P\int_0^\pi \frac{\tan x}{x}dx + P\int_0^\pi\frac{\tan x}{\pi + x}dx + P\int_0^\pi \frac{\tan x}{2\pi + x}dx+\ldots$$ and then we have $$ P\int_0^\pi \frac{\tan x}{k\pi +x}dx = \int_0^{\pi/2} \tan x\left(\frac{1}{k\pi+x} - \frac{1}{(k+1)\pi-x}\right)dx $$ Finally, $$ \sum_{k=0}^\infty \frac{1}{k\pi+x} - \frac{1}{(k+1)\pi-x} =\cot x,$$ so $$ P\int_0^\infty \frac{\tan{x}}{x}dx = \int_{0}^{\pi/2}\tan x\cot x \;dx = \pi/2.$$

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  • $\begingroup$ what is the variable P? $\endgroup$ – IntegrateThis Jan 13 '18 at 8:24
  • $\begingroup$ @IntegrateThis Not a variable. A notation for Cauchy principal value. $\endgroup$ – spaceisdarkgreen Jan 13 '18 at 8:27
  • $\begingroup$ would $ \int_0^{\infty-1}\frac{\tan x}{x}\mathrm dx$ virtually be $\pi/2$? Would be the $o(x)$ error given by $1/x$ which would be near zero by the time $x=\infty-1$? $\endgroup$ – onepound Dec 30 '19 at 16:14
  • $\begingroup$ @onepound what does $\infty-1$ mean? $\endgroup$ – spaceisdarkgreen Dec 30 '19 at 17:30
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This can also be handled using contour integration. Since there are no singularities inside $\gamma$, $$ \int_\gamma\frac{\tan(z)}z\,\mathrm{d}z=0 $$ where $\gamma$ is the contour

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where $k\to\infty$ then $m\to\infty$.

First notice that the integrals along all the small arcs, which are $-\pi i$ times the residues at the singularities of $\frac{\tan(z)}z$, cancel out for each $k$.

Since $\tan(z)=i\tanh(\operatorname{Im}(z))$ for $\operatorname{Re}(z)\in\mathbb{Z}$, the absolute value of the integrals along the vertical lines are at most $\frac mk$. These vanish as $k\to\infty$.

Therefore, we get $$ \mathrm{PV}\int_{-\infty}^\infty\frac{\tan(x)}x\,\mathrm{d}x =\int_{-\infty+mi}^{\infty+mi}\frac{\tan(z)}z\,\mathrm{d}z $$ as $m\to\infty$, $\tan(z)\to i$. Therefore, $$ \mathrm{PV}\int_{-\infty}^\infty\frac{\tan(x)}x\,\mathrm{d}x=\pi $$ Since $\frac{\tan(x)}x$ is even, we get that $$ \mathrm{PV}\int_0^\infty\frac{\tan(x)}x\,\mathrm{d}x=\frac\pi2 $$

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