1

What is the mistake in the calculation of $$\int_0^\infty {\cos(ax)\over(x^2+1)^2}dx$$ where $0<a\in\mathbb{R}$?

I know that the question itself has already an answered in here but when I comment about questions I rarely get answers so I opned a new question, I hope that ok, sorry in advence if it's not.

My worng attempt: $$ I:= \int_0^\infty {\cos(ax)\over(x^2+1)^2}={1\over2}\int_{-\infty}^\infty{\cos(ax)\over(x^2+1)^2}= \\ \int_{C_R}{\cos(ax)\over(x^2+1)^2} $$ where $C_R$ is the half semicircle centered at the origin with radii $R$ joint with $[-R,R]$.

The above equality holds because $\int_{-R}^R=0$. $$I= {1\over2}\cdot 2\pi i\text{Res}({\cos(ax)\over(x^2+1)^2},i)=\\ \pi i({\cos(ax)\over(x^2+i)^2})'|_{z=i}=\\ \pi i\cdot({-a\sin(az)(z+i)^2-2(z+i)\cos(az)\over (z+i)^4})|_{z=i} =\\ {\pi i\over 16}(a\sin(ai)\cdot 4-4i\cos(ia)=\\ {\pi\over 4}(\cos(ai)+ia\sin(ai))={\pi\over 8}(e^{ai}+e^{-ai}+a(e^{ai}-e^{-ai}))={\pi\over8}(e^{ai}(1+a)+e^{-ai}(1-a)) $$ But the answer should be ${\pi\over 2e}$.

Community
  • 1
J. Doe
  • 1,336
  • Note that your answer must be wrong, because it has imaginary parts, where the function is only integrated along the real line. – The Count Jul 03 '19 at 20:28

1 Answers1

3

$I\neq\displaystyle\lim_{R\to\infty}\int_{C_R}\frac{\cos ax\,dx}{(x^2+1)^2}$ because the limit over the semicircle doesn't exist (as $\cos ax$ is unbounded there; you have to use $e^{iax}$ for $a>0$ instead).

UPDATE: As noted by @Maxim, the limit still exists (I should have noted it myself as it follows from your evaluation). But the rest of above applies.

metamorphy
  • 39,111