1
$\begingroup$

I want to show that (for $x \in \mathbb{R}$ and $z \in \mathbb{C}$)

$$ \int_{-\infty}^{\infty} {\dfrac{\cos \pi x}{z^2-2z+5}}\mathrm{d}x = -\frac{\pi }{2}e^{-2\pi}$$

However, I am a little confused on how to proceed given that the denominator of the integrand is complex.

If the integral was in the form $ \int_{-\infty}^{\infty} {\dfrac{\cos \pi x}{\color{blue}{x^2-2x+5}}}\mathrm{d}x$, then I could treat it as a Fourier integral, and use the fact that $$ \int_{-\infty}^{\infty} {\dfrac{\cos \pi x}{x^2-2x+5}}\mathrm{d}x = -2\pi \sum {\operatorname{Im} \operatorname{Res} \left[ \dfrac{e^{i\pi z}}{z^2-2z+5} \right] }$$

But given that the numerator of the integrand is a complex number, I am missing a key insight on how to proceed towards a solution. Thanks in advance!

$\endgroup$
3
$\begingroup$

Let $\Gamma_{R}$ be the semicircular contour in the upper-half plane of radius $R$. Then calculate

$$\int_{-\infty}^{\infty} \frac{\cos(\pi x)}{x^2-2x+5}\, dx = \Re \left \{\int_{-\infty}^{\infty} \frac{e^{i \pi x}}{x^2-2x+5}\, dx \right \} \\ = \Re \left \{\int_{\Gamma_R} \frac{e^{i \pi z}}{z^2-2z+5} \right \} \\ = \Re \{2\pi i \mathrm{Res}_{z=1+2i} \left (e^{i \pi z}/(z^2-2z+5) \right ) \} \\ = \Re \{2\pi i e^{i \pi - 2\pi}/(4i) \} = -\frac{\pi}{2}e^{-2\pi}.$$

On the question of whether it is a $z$ or an $x$: it is a typo. Just substitute $z=0$ and $z=1$ to belie a contradiction by linearity to do a dummy check.

$\endgroup$

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.