This question has an answer that relates differentiation under the integral to the OP.

Again, here's the original integral: $$\int_0^\infty\frac{\cos\;x}{1+x^2}\mathrm{d}x$$

...and we let $$ F(y) = \int\nolimits_{0}^{\infty} \frac{\sin xy}{x(1+x^2)} \ dx \ \ \text{for} \quad\quad y > 0$$

The first part of interest is in showing that $\displaystyle F''(y) - F(y) + \pi/2 = 0$. Is it necessary to integrate $F(y)$ to show this? What about the possibility of taking $\lim_{y \to 0+}$ beforehand? I'm wondering if someone can help explain this step in much greater detail. I'm a little hazy with the $y>0$ portion of it, and whether or not integration has to occur here. I'm trying to make sure I thoroughly understand this post so that I can apply it later to different problems.


Computing the following $F''(y) - F(y)$ combination, gives

$$ \int_0^\infty \frac{\sin(x y)}{x} \frac{-x^2 -1 }{1+x^2} \mathrm{d} x = -\int_0^\infty \sin (x y) \frac{\mathrm{d} x}{x} = -\int_0^\infty \sin (x) \frac{\mathrm{d} x}{x} = -\frac{\pi}{2} $$ Notice that we did use $y>0$ when changing variables.

Now, the only thing that remains to be justified is that $F''(y)$ integral converges, and this is so because $\frac{x^2}{1+x^2} < 1$ for real $x$.

  • $\begingroup$ I'm really sorry for missing something so basic, but how did you go from $\sin{xy}$ to $\sin{x}$? $\endgroup$ – Matt Groff Aug 27 '11 at 22:54
  • $\begingroup$ He let $u=xy$ then changed back to $x$'s. $\endgroup$ – Ragib Zaman Aug 28 '11 at 1:53
  • $\begingroup$ @Ragib Zaman: Thanks. :-) $\endgroup$ – Matt Groff Aug 28 '11 at 17:13

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