1
$\begingroup$

I wanted to evaluate, using residue theorem: \begin{align*} \int_{-\infty}^{+\infty} \frac{\cos(mx)}{x^2+1} \end{align*}

Then I use the semicircumference $(C_R)$ of radius $R>2$ and center in $0$ and $f(z)=\dfrac{e^{i(mz)}}{z^2+1}$ because it only has a singularity in $C_R$ $(i)$ and $\Re(f)= \dfrac{\cos{mz}}{z^2+1}$. Then I get: \begin{align*} \oint_{C_R} f(z)dz = \int_{[-R,R]} f(z)dz + \int_{γ_R} f(z)dz $ \end{align*} \begin{align*} Res(f,i) = \lim_{z\to i} \frac{e^{i(mz)}}{z+i} = \frac{e^{-m}}{2i} = -\frac{i}{2e^m} \end{align*} Hence:

\begin{align*}\oint_{C_R} f(z)dz = 2\pi i (-\frac{i}{2 e^m}) = \frac{\pi}{e^m}\end{align*}

Moreover:

\begin{align*} \left|\int_{γ_R} f(z)dz\right| & \leq \int_{γ_R} \frac{|e^{i(mz)}|}{|z^2+1|} dz \leq \int_{γ_R} \frac{|e^{i(x+iy)}|}{|z^2|-1} dz = \int_{γ_R} \frac{|e^{i(x+iy)}|}{|z^2|-1} dz = \int_{γ_R} \frac{|e^{-y+ix}|}{R^2-1} dz\\ & = \int_{γ_R} \frac{e^{-y}}{R^2-1} dz \leq \int_{γ_R} \frac{1}{R^2-1} dz \leq \frac{\pi R}{R^2-1} \xrightarrow{R\to\infty} 0 \end{align*}

And supposedly I win. But, my question is: Why cannot I use $f(z)=\dfrac{\cos{mz}}{z^2+1}$ and do the same reasoning?

  • 2
    using $\cos$ directly you will get into troubles into finding a suitable contour of integration. the half circle won't do it.. – 2017-01-02
  • 0
    @tired Sorry, man, I don't understand why I get into troubles with the half circle . – 2017-01-02
  • 1
    how does $\cos(m z)$ behaves along the imaginary axis? – 2017-01-02
  • 0
    @tired I feel like 5 years old now. Thank you. – 2017-01-02

1 Answers 1

1

The function $\cos(mz)$ does not fullfill Jordan's lemma on the upper half plane, while $e^{miz}$ (assuming $m>0$) does, so it is better to tackle the problem this way:

$$ \int_{-\infty}^{+\infty}\frac{\cos(mz)}{z^2+1}\,dz = \text{Re}\int_{-\infty}^{+\infty}\frac{e^{miz}}{z^2+1}\,dz = \text{Re}\left(2\pi i\, \text{Res}\left(\frac{e^{miz}}{z^2+1},z=i\right)\right)\tag{1} $$ and since $z=i$ is a simple pole for $\frac{e^{miz}}{z^2+1}$, the residue equals $\frac{e^{-m}}{2i}$ and the integral equals $\color{red}{\large\frac{\pi}{e^m}}$.