I am looking to solve $$I=\int_{-\infty}^{\infty}\frac{\cos{x}}{x^2+1}dx$$ To do so i defined a function of complex variable as follows: $$f(z)=\frac{\cos{z}}{z^2+1}$$ Then i make a closed countour $C$ by uniting a semicircle (denoted $\gamma_R$) above the origin and a line connecting points $-R$ and $R$ on the real axis. $$\oint_Cf(z)dz=\int_{-R}^{R}f(x)dx+\int_{\gamma_R}f(z)dz$$ The only pole within the contour is at $z=i$. By the residue theorem $$\oint_Cf(z)dz=2\pi i\cdot \operatorname{Res}[f]_{z=i}$$ I use the definition of complex cosine: $$\cos{z}=\frac{e^{iz}+e^{-iz}}{2}$$ $$\operatorname{Res}[f]_{z=i}=\lim_{z\to i}\bigg\{(z-i)\frac{e^{iz}+e^{-iz}}{2(z-i)(z+i)}\bigg\}=\frac{1/e+e}{4i}$$ My calculation then becomes: $$I=\pi\frac{1/e+e}{2}-\int_{\gamma_R}f(z)dz$$ Now: $$\int_{\gamma_R}f(z)dz=\int_{-R}^{R}f(Re^{i\phi})dRe^{i\phi}=\int_{-R}^{R}\frac{\cos{(Re^{i\phi})}}{R^2e^{2i\phi}+1}d\phi$$ as $R\to\infty$ this integral goes to zero (Jordan's Lemma??). I can conclude $$I=\pi\frac{e+1/e}{2}$$ But the answer should be $$\frac{\pi}{e}$$ I can't seem to find the mistake. I am pretty new to complex analysis, would anyone give me hint, what did i do wrong please?


The reason your method does not work is that the integral over $\gamma_R$ does not tend to zero as $R$ tends to infinity because $e^{-iz}$ blows up.

There is a standard trick here. Consider $\cos(\theta) = \operatorname{Re}(e^{i\theta})$ and integrate $\frac{e^{iz}}{1+z^2}$ instead. The integral of this over $\gamma_R$ is zero because $e^{iz}$ is bounded.

  • $\begingroup$ Ah, yea, you're right, didnt realise that the cosine function is not bounded by $(-1,1)$ in the complex plane. When i rewrite the cosine function the numerator is $$e^{iRe^{i\phi}}+e^{-iRe^{i\phi}}=e^{...}+e^{-iR(cos\phi+i\sin{\phi})}$$ and the $e^{R\sin{\phi}}$ part blows up, i see it now. $\endgroup$ – Michal Dvořák Apr 2 '18 at 13:38
  • $\begingroup$ You might find it helpful to google Louville's theorem - in fact no everywhere differentiable functions are bounded on $\mathbb{C}$ $\endgroup$ – Patrick Apr 2 '18 at 13:41

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.