Evaluate integral $\int_{-\infty}^{+\infty} \frac{\sin{(x - 3)}}{x - 3} e^{-|x|} dx$

Question

Evaluate the integral $\int_{-\infty}^{+\infty} \frac{\sin{(x - 3)}}{x - 3} e^{-|x|} dx$. I tried to use Fourier transforms of functions $\frac{\sin{x}}{x}$ and $e^{-|x|}$, because if $f(x) = \frac{\sin{(x - 3)}}{x - 3}$ and $g(x) = e^{-|x|}$, then

$$ \int_{-\infty}^{+\infty} \frac{\sin{(x - 3)}}{x - 3} e^{-|x|} dx = \widehat{f \cdot g}(0) = [\hat{f} * \hat{g}](0) $$

But i ended up with some other weird integral $$ \int_{-\infty}^{+\infty} \frac{\sin{(x - 3)}}{x - 3} e^{-|x|} dx = \int_{-1}^{1} \frac{e^{3ix}}{1+x^2}dx $$

Answer 1

Your integral can be transformed into a "less weird" one :

$$I= 2\int_{0}^{1} \frac{\cos(3x)}{1+x^2}dx$$

by splitting the integration on $[-1,1]$ into two integrals, one on $[-1,0]$, the other one on $[0,1]$, then making the change of variables $x=-X$ in the first one, and using Euler formula

$$2 \cos(a)=e^{ia}+e^{-ia}$$ One obtains numerical value $\approx 0.294769$ but it does not seem there is an exact value with usual functions.

Answer 2

$$I(t)=\int_{-\infty}^\infty\frac{\sin(t(x-3))}{x-3}e^{-|x|}dx=\int_{-\infty}^\infty\frac{\sin(tu)}{u}e^{-|u+3|}du$$ $$I'(t)=\int_{-\infty}^\infty\cos(tu)e^{-|u+3|}du=\frac{2\cos(3t)}{t^2+1}$$ $$I(t)=\int\frac{2\cos(3t)}{t^2+1}dt$$ Now since we want $I=I(1)$ we can say: $$I=\int_0^1\frac{2\cos(3t)}{t^2+1}dt$$

Answer 3

Starting from the integral given in other answers, we can write $$I= 2\int_{0}^{1} \frac{\cos(3x)}{1+x^2}\,dx=3i\left(\int_{0}^{1}\frac{\cos(3x+3i-3i)}{ 3x+3i}\,dx-\int_{0}^{1}\frac{\cos(3x-3i+3i)}{3 x-3i}\,dx\right)$$ after clear substitutions and expansions of complex cosines, we end with $$I=(\text{Si}(-3+3 i)-\text{Si}(3+3 i)) \sinh (3)-i (\text{Ci}(-3+3 i)-\text{Ci}(3+3 i)) \cosh (3)$$ which does not seem possible to simplify.

Its numerical value is $0.29476885324358497714$