Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm doing a self-study in learning contour integration in complex analysis and here's an example i came across:

$$\int_{-\infty}^{\infty} \frac{\cos x + x \sin x}{1+x^2} dx$$

How does this relate to the following integral over a semi-circle in the upper half plane ($0\leq t \leq \pi$)?

$$\int_{R e^{it}} \frac{e^{iz}}{z-i} dz$$

(This came from a hint in our notes)

share|improve this question

1 Answer 1

Note that $$ f(z) = \frac{e^{iz}}{z-i} = \frac{(z+i)e^{iz}}{(z+i)(z-i)} = \frac{(z+i)e^{iz}}{z^2+1}. $$

So, if $z=x$ is real, then $$ f(x) = \frac{(x+i)(\cos x + i \sin x)}{x^2+1} = \frac{x\cos x - \sin x + i(\cos x + x\sin x)}{x^2+1}. $$

Let $\Gamma$ be the closed curve consisting of the interval $[-R,R]$ together with the semi-circle $C : z=Re^{it}$, $0 < t < \pi$ (for $R > 1$). By the residue theorem,

$$ \int_{\Gamma} f(z)\,dz = \int_{[-R,R]} f(z)\,dz + \int_C f(z)\,dz = 2\pi i \operatorname{Res}(f;z=i) $$

(since $z=i$ is the only singularity of $f$ inside $\Gamma$). Finally, let $R\to\infty$.

The integral over $C$ will tend to $0$ by Jordan's lemma, and you will end up with

$$ \int_{-\infty}^\infty \frac{x\cos x - \sin x}{x^2+1}\,dx + i \int_{-\infty}^\infty \frac{x\sin x + \cos x}{x^2+1}\,dx = 2\pi i \operatorname{Res}(f;z=i). $$

To finish off, compute the residue and look at the imaginary part of the equality above.

share|improve this answer
Just a caution: Since a priori we don't know whether this integral, we indeed need to choose a slight different contour, a rectangle on the upper half plane, for details, see P 156, 3) in the book "complex analysis" by Ahlfors, which deals with the case with simple zero at infinity. –  Lao-tzu Jan 12 at 8:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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