4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Substitute $u = x\csc(x)$. $\endgroup$ Jul 21 '19 at 2:30
  • $\begingroup$ Are you sure it converges? $\endgroup$ Jul 21 '19 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 '19 at 2:55
16
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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 '19 at 4:03
  • $\begingroup$ @robyjohn, Thanks for point it our. I have edited it now. $\endgroup$
    – Z Ahmed
    Jul 21 '19 at 4:27
5
$\begingroup$

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 $$ \begin{align} \int_0^\infty\frac{\sin(x)-x\cos(x)}{x^2+\sin^2(x)}\,\mathrm{d}x &=-\frac1{2i}\log\left(\frac{1-i}{1+i}\right)\\ &=\frac\pi4 \end{align} $$

$\endgroup$
1
  • 1
    $\begingroup$ $\frac1{2i}\log\left(\frac{x-i\sin(x)}{x+i\sin(x)}\right) =-\tan^{-1}\left(\frac{\sin(x)}x\right)$ $\endgroup$
    – robjohn
    Jul 21 '19 at 4:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.