Calculate $\displaystyle \frac{1}{\sqrt{2\pi}} \int^\infty_{-\infty} \frac{e^{-ikx}}{1 + k^2} dk$.
How should I start?
You should start considering that is...
$\displaystyle \int_{- \infty}^{+\infty} \frac{e^{-j k x}}{1+k^{2}}\ dk = \int_{- \infty}^{+\infty} \frac{e^{j k x}}{1+k^{2}}\ dk $ (1)
... and then compute the integral (1) integrating the complex variable function...
$\displaystyle f(z)= \frac{e^{j z x}}{1+z^{2}}$ (2)
... along the path $\displaystyle \Gamma$ in the figure...
... if R tends to infinity. For doing that the Cauchy integral formula may be useful...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Thank you. I get it now. The contour integral over $\displaystyle \Gamma$ which does not enclose a pole will be 0 according to Cauchy integral theorem (Cauchy Integral Formula -- from Wolfram MathWorld). Hence the countour integral over $\displaystyle [-R, R]$ is equal to the countour integral of the semicircle in red ($\displaystyle \Gamma$ and $\displaystyle [-R,R]$). Then we can use either the Cauchy integral formula or the Residue formulas to calculate the semicircle countour integral. The Residue formula could be more convenient sometimes as we won't have to do partial fractions/complete square that we sometimes need to do when using Cauchy integral formula approach.
Thanks very much for helping me. (I didn't realize this is a complex integration and was trying to solve it as a real integral lol.)
The formula I spoke about gives the integral of an f(z) along a closed path $\displaystyle \Gamma$...
$\displaystyle \int_{\Gamma} f(z)\ dz = 2\ \pi\ j\ \sum_{i} r_{i}$ (1)
... where the $\displaystyle r_{i}$ are the residues of the poles of f(z) inside $\displaystyle \Gamma$. In Your case the only pole of f(z) inside $\displaystyle \Gamma$ is at z=j [see figure]...
... and the residue is...
$\displaystyle r= \lim_{z \rightarrow j} (z-j)\ \frac{e^{j z x}}{1+z^{2}}= \frac{e^{-x}}{2j}$ (2)
If R tends to infinity the integral along the 'bif half circle' tends to 0 so that the result is...
$\displaystyle \int_{-\infty}^{+\infty} \frac{e^{j k x}}{1+k^{2}}\ dk = \pi\ e^{-x}$ (3)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Thanks for writing out the solutions. I calculated without changing the sign in the beginning, i.e. I evaluated the LHS of below equation that you wrote.
$\displaystyle \int_{- \infty}^{+\infty} \frac{e^{-j k x}}{1+k^{2}}\ dk = \int_{- \infty}^{+\infty} \frac{e^{j k x}}{1+k^{2}}\ dk (1)$
The solution I got for the last step is $\displaystyle \pi e^{x}$ instead of $\displaystyle \pi e^{-x}$.
Why does equation (1) hold? How can the LHS and RHS yield different results if they are equal?
Because $\displaystyle e^{j k x}$ and $\displaystyle e^{-j k x}$ have the same real part and the inaginary part of Your integral is 0 the substituion is perfectly possible... but why the substitution?... the reason is that using $\displaystyle e^{j z}$ the value of the function on the 'upper big half circle' is $\displaystyle e^{j R\ e^{j \theta}}= e^{j R\ \cos \theta}\ e^{-R \sin \theta}$ and it tends to 0 if R tends to infinity, so that the Jordan's lemma is satisfied...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$