I dont know how to evaluate this improper integral: $$I=\int_{-1}^1 \left(\frac{1}{x}+1-\frac{\sqrt{1-x^2}}{x}\right )\arctan\frac{2}{x^2}dx$$ At first I tried to do it in trigonometric substitution：$x=\sin t,dx=\cos tdt$ $$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\csc t+1-\cot t\right)\arctan(2\csc ^2t)\cos tdt$$ $$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\cot t+\cos t-\cot t\cos t\right)\arctan(2\csc ^2t)dt$$ I don't know what to do next,any help will be appreciated.
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See https://math.stackexchange.com/questions/2427602/find-the-value-of-infinite-series-sum-n-1-infty-tan-12-n2#comment5014314_2427602 – lab bhattacharjee Sep 15 '17 at 05:14
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Thank you for the link, but that is a sum,but does that have anything to do with the integral? – JamesJ Sep 15 '17 at 05:19
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Write $$\arctan\dfrac2{x^2}=\arctan\dfrac1{x-1}-\arctan\dfrac1{x+1}$$ – lab bhattacharjee Sep 15 '17 at 05:22
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@,Thanks for your hint,I'll try trying. – JamesJ Sep 15 '17 at 05:38
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$\left(\frac{1}{x}-\frac{\sqrt{1-x^2}}{x}\right )\arctan\frac{2}{x^2}$ is an odd function and it's continue at $x=0$ – FDP Sep 15 '17 at 12:54
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@FDP,Yes I see,thanks a lot ,is there a similar amazing step next? – JamesJ Sep 15 '17 at 14:23
2 Answers
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Hint:
$$\begin{align} \mathcal{I} &=\int_{-1}^{1}\mathrm{d}x\,\left(\frac{1}{x}+1-\frac{\sqrt{1-x^{2}}}{x}\right)\arctan{\left(\frac{2}{x^{2}}\right)}\\ &=\int_{-1}^{0}\mathrm{d}x\,\left(\frac{1}{x}+1-\frac{\sqrt{1-x^{2}}}{x}\right)\arctan{\left(\frac{2}{x^{2}}\right)}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\left(\frac{1}{x}+1-\frac{\sqrt{1-x^{2}}}{x}\right)\arctan{\left(\frac{2}{x^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\left(-\frac{1}{x}+1+\frac{\sqrt{1-x^{2}}}{x}\right)\arctan{\left(\frac{2}{x^{2}}\right)};~~~\small{\left[x\mapsto-x\right]}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\left(\frac{1}{x}+1-\frac{\sqrt{1-x^{2}}}{x}\right)\arctan{\left(\frac{2}{x^{2}}\right)}\\ &=2\int_{0}^{1}\mathrm{d}x\,\arctan{\left(\frac{2}{x^{2}}\right)}.\\ \end{align}$$
In other words, only the even components of the integrand contribute to value of the integral because of the symmetry of the limits of integration.
David H
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,Wow,beautifull answer,great! thank you very much, and then what to do next? – JamesJ Sep 15 '17 at 09:46
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@JamesJ Integration by parts is one possibility. That gives you an integral of a simple rational function, so the final answer will be in terms of elementary functions. – David H Sep 15 '17 at 09:57
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Let me write a complete answer： \begin{align} I&=\int_{-1}^1\left(\frac{1}{x}+1-\frac{\sqrt{1-x^2}}{x}\right)\arctan\left(\frac{2}{x^2}\right)dx\\ &=\int_{-1}^1\arctan\left(\frac{2}{x^2}\right)dx\qquad(\text odd function)\\ &=\int_0^2\arctan\frac{2}{(x-1)^2}dx\\ &=\int_0^2\arctan\frac{x+(2-x)}{1-x(2-x)}dx\\ &=\int_0^2(\arctan x+\arctan(2-x))dx\\ &=2\int_0^2\arctan xdx\\ &=2(x\arctan x|_0^2-\int_0^2\frac{x}{1+x^2}dx)\\ &=4\arctan 2-\ln (1+x^2)|_0^2\\ &=4\arctan 2-\ln 5 \end{align}
JamesJ
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