I want to evaluate

$$\int_{0}^{\infty}\frac{\sin(\pi x)}{1 + x^2}$$

by the residue theorem.

This function is not even, so I can not use the semi-circle approach.

I was trying to integrate over the triangle $T= \{(0,0),(T,0),(T,T)\}$ and then take the limit as $T\to\infty$ but I had no luck with the integrals.

I parametrize each edge of the triangle and I know that the function is analytic on T and the contour integral is 0. I dont know how to deal with the remaining integrals over the sides of T.

  • 1
    $\begingroup$ It never hurts to replace trig functions with $e^{ix}$ $\endgroup$ – Simply Beautiful Art Aug 12 '17 at 0:07
  • $\begingroup$ @Hans The integral here is not over the entire real line though. $\endgroup$ – Simply Beautiful Art Aug 12 '17 at 0:11
  • $\begingroup$ @Hans I know about that substitution and I have used that before, but the sine function is not even, so we can not use the semi-circle approach with my function also because the integral is from $0$ to $\infty$ if it is possible, please show me how. $\endgroup$ – Richard Clare Aug 12 '17 at 0:21
  • $\begingroup$ @SimplyBeautifulArt and RichardClare, you are both right. My bad. I have deleted my false comment. Let use delete the corresponding comments. $\endgroup$ – Hans Aug 12 '17 at 0:22
  • $\begingroup$ @Hans don't worry and thank you anyway. $\endgroup$ – Richard Clare Aug 12 '17 at 0:23

Without contour integration.

As Simply Beautiful Art commented, writing

$$\frac{\sin(\pi x)}{1+x^2}=\frac{\sin(\pi x)}{(x+i)(x-i)}=\frac i2 \left(\frac{\sin(\pi x)}{x+i}- \frac{\sin(\pi x)}{x-i}\right)$$ and using $$\int \frac{\sin (a x)}{x+b} \,dx=\cos (a b)\, \text{Si}(a(x+ b))-\sin (a b)\, \text{Ci}(a(x+ b))$$ $$\int_0^\infty \frac{\sin (a x)}{x+b} \,dx=\frac{a (2 \text{Ci}(b |a|) \sin (b |a|)+\cos (a b) (\pi -2 \text{Si}(b |a|)))}{2 |a|}$$ we should end with $$\int_0^\infty \frac{\sin(\pi x)}{1+x^2}\,dx=\text{Shi}(\pi ) \cosh (\pi )-\text{Chi}(\pi ) \sinh (\pi )$$ where appear the hyperbolic sine and cosine integral functions.

  • $\begingroup$ Sure, this is essentially the definition of $\mathrm{Si}$ and $\text{Ci}$ $\endgroup$ – Hans Aug 14 '17 at 4:12

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