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The integral $$\int_{0}^{\infty} \frac{\sin x-x\cos x}{x^2+\sin^2x } dx$$ admits a nice closed form. The question is: How to evaluate it by hand.

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  • $\begingroup$ Substitute $u = x\csc(x)$. $\endgroup$ Jul 21, 2019 at 2:30
  • $\begingroup$ Are you sure it converges? $\endgroup$ Jul 21, 2019 at 2:36
  • $\begingroup$ The integral is imroper but it is convergent, one has to take limit of the anti-derivative as $x \rightarrow \infty$. $\endgroup$
    – Z Ahmed
    Jul 21, 2019 at 2:55

2 Answers 2

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$$ I=\int_{0}^{\infty} \frac{\sin x-x \cos x}{x^2+\sin^2 x} dx= - \int_{0}^{\infty}\frac{\frac {x\cos x -\sin x}{x^2}}{1+(\frac{\sin x}{x})^2}dx= -\int_{1}^{0} \frac{dt}{1+t^2}=\frac{\pi}{4}.$$

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  • $\begingroup$ Isn't $-\int_0^\infty\frac{\mathrm{d}t}{1+t^2}=-\frac\pi2$? The integral should be $-\int_1^0\frac{\mathrm{d}t}{1+t^2}$. $\endgroup$
    – robjohn
    Jul 21, 2019 at 4:03
  • $\begingroup$ @robyjohn, Thanks for point it our. I have edited it now. $\endgroup$
    – Z Ahmed
    Jul 21, 2019 at 4:27
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Note that $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\log(x-i\sin(x)) &=\frac{1-i\cos(x)}{x-i\sin(x)}\\ &=\frac{x+\sin(x)\cos(x)}{x^2+\sin^2(x)}+i\frac{\sin(x)-x\cos(x)}{x^2+\sin^2(x)} \end{align} $$ Taking the imaginary part of both sides $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\frac1{2i}\log\left(\frac{x-i\sin(x)}{x+i\sin(x)}\right) &=\frac{\sin(x)-x\cos(x)}{x^2+\sin^2(x)} \end{align} $$ Thus, $$ \int\frac{\sin(x)-x\cos(x)}{x^2+\sin^2(x)}\,\mathrm{d}x =\frac1{2i}\log\left(\frac{x-i\sin(x)}{x+i\sin(x)}\right)+C $$ and $$ \begin{align} \int_0^\infty\frac{\sin(x)-x\cos(x)}{x^2+\sin^2(x)}\,\mathrm{d}x &=-\frac1{2i}\log\left(\frac{1-i}{1+i}\right)\\ &=\frac\pi4 \end{align} $$

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    $\begingroup$ $\frac1{2i}\log\left(\frac{x-i\sin(x)}{x+i\sin(x)}\right) =-\tan^{-1}\left(\frac{\sin(x)}x\right)$ $\endgroup$
    – robjohn
    Jul 21, 2019 at 4:27

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