Question

The value of the integral ${\int }_{\infty }^{-\infty }\frac{sinx}{x^2+2x+2}dx$ evaluated using contour integration and the residue theorem is

• A. $-\pi sin\left(1\right)/e$
• B. $-\pi cos\left(1\right)/e$
• C. $sin\left(1\right)/e$
• D. $cos\left(1\right)/e$

Answer 1

The correct option is A $-\pi sin\left(1\right)/e$

Given:
$I={\int }_{\infty }^{-\infty }\frac{sinx}{x^2+2x+2}dx$
Now, Let us assume:
$e^{jx}=cosx+jsinx$
$\Rightarrow sinx=Imag\left[e^{jx}\right];$
Now, $I={\int }_{-\infty }^{\infty }\frac{Imag\left[e^{jx}\right]}{x^2+2x+2}dx;$
Let us assume a new function
$f\left(z\right)=\frac{Imag\left[e^{iz}\right]}{z^2+2z+2}$
Where
$z^2+2z+2=0$
represents poles of $f\left(z\right)$
$\Rightarrow z=\frac{-2\pm \sqrt{4-4\left(2\right)}}{2}$
$=\frac{-2\pm \sqrt{-4}}{2}=\frac{-2\pm j2}{2}$
$z=-1\pm j1$
Form the $z-$plane it is clear that $z=-1+j1$ is the only pole lying in $f\left(z\right)>0,$ ie.e., in upper half plane.



$\underset{z\to -1+j1}{Resf\left(z\right)}=\underset{z\to -1+jq}{lim}[z-(-1+j1\left)\right]\times$
$\frac{e^{jz}}{[z-(-1+j1\left)\right][z-(-1-j1\left)\right]}$
$=\underset{z\to 1+j1}{lim}\frac{e^{jz}}{z+(1+j1)}$$=\frac{e^{j(-1+j1)}}{-1+j1+1+j1}=\frac{e^{-j-1}}{j2}$
$\therefore {\int }_c\frac{Imag\left(e^{jz}\right)}{z^2+2z+2}dz=Imag[j2\pi .\frac{e^{-j-1}}{j2}]$
(using Residue theorem)
$=Imag[j2\pi .\frac{e^{-j}}{j2e}]$
$=Imag\left[\frac{\pi e^{-j}}{e}\right]$
$=Imag\left[\frac{\pi }{e}\right[cos\left(1\right)-jsin\left(1\right)\left]\right]$
$=-\frac{\pi sin\left(1\right)}{3}$
[taking imaginary parts]