# Evaluate $I=\int\limits_{-\infty}^{+\infty}\frac{\sin(x-3)}{x-3}e^{-|x|}\ dx$

How to evaluate this integral? $$I=\int\limits_{-\infty}^{+\infty}\frac{\sin(x-3)}{x-3}e^{-|x|}\ dx$$

I tried to use the properties of Fourier transform $$\left(\text{in the form of }\hat{f}(y)=\int\limits_{-\infty}^{+\infty}f(x)e^{ixy}dx\right)$$: $$f(x)=\frac{\sin(x-3)}{x-3},\ \ g(x)=e^{-|x|}\\ I=\int\limits_{-\infty}^{+\infty}f(x)g(x)\ dx=\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}\hat{f}(y)\hat{g}(y)\ dy=\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}e^{3iy}\pi I_{[-1,1]}(y)\cdot\frac{2}{1+y^2}\ dy=\\ =\int\limits_{-1}^1\frac{e^{3iy}}{1+y^2}\ dy=2\int\limits_0^1\frac{\cos (3y)}{1+y^2}\ dy=\ ?$$

And the last integral doesn't seem to be solvable. I tried a few other ways of solving this problem via Fourier transform, and all of them led to this integral.

So, now, I am questioning myself if there is a solution to $$I$$ in terms of elementary functions.

• I believe you can compute this using contour integration and some complex analysis Mar 9, 2020 at 22:41

Concerning the antiderivative $$I=2\int\frac{\cos (3y)}{1+y^2}\, dy=i\left(\int \frac{\cos (3y)}{y+i}\ dy-\int \frac{\cos (3y)}{y-i}\ dy\right)$$ that is to say $$I=3i\left(\int \frac{\cos (3y+3i-3i)}{3y+3i}\ dy-\int \frac{\cos (3y-3i+3i)}{3y-3i}\ dy\right)$$ I suppose that the changes of variable are clear and this makes $$I=\sinh (3) (\text{Si}(3 i-3 y)-\text{Si}(3 y+3 i))-i \cosh (3) (\text{Ci}(3 i-3 y)-\text{Ci}(3 y+3 i))$$
So, $$J=2\int_0^1\frac{\cos (3y)}{1+y^2}\, dy=(\text{Si}(-3+3 i)-\text{Si}(3+3 i)) \sinh (3)-i (\text{Ci}(-3+3 i)-\text{Ci}(3+3 i)) \cosh (3)$$ which is $$0.294768853243585$$. This number is not identified by inverse symbolic calculators.
Let $$f(x)$$ be analytic in $$\textbf{R}$$. Then $$f(x)=\sum^{\infty}_{k=0}\frac{f^{(k)}(0)}{k!}x^k$$ Hence $$I(a)=\int^{\infty}_{-\infty}f(t-a)e^{-|t|}dt=\int^{\infty}_{0}f(t-a)e^{-t}dt+\int^{\infty}_{0}f(-t-a)e^{-t}dt=$$ $$=\int^{\infty}_{0}\sum^{\infty}_{k=0}\frac{f^{(k)}(0)}{k!}(t-a)^ke^{-t}dt+\int^{\infty}_{0}\sum^{\infty}_{k=0}\frac{f^{(k)}(0)}{k!}(-t-a)^ke^{-t}dt=$$ $$=\sum^{\infty}_{k=0}\frac{f^{(k)}(0)}{k!}\int^{\infty}_{0}(t-a)^ke^{-t}dt+\sum^{\infty}_{k=0}\frac{f^{(k)}(0)}{k!}(-1)^k\int^{\infty}_{0}(t+a)^ke^{-t}dt=$$ $$=\sum^{\infty}_{k=0}\frac{f^{(k)}(0)}{k!}C_{k}(a),$$ where $$C_k(a)=e^{-a}\Gamma(k+1,-a)+(-1)^k e^a\Gamma(k+1,a)$$ Here $$f(x)=\frac{\sin(x)}{x}$$ is even function of $$x$$. Hence $$I(a)=\sum^{\infty}_{k=0}\frac{(-1)^k}{(2k+1)!}\left(e^{-a}\Gamma(2k+1,-a)+e^{a}\Gamma(2k+1,a)\right)$$ But $$\Gamma(2k+1,a)=e^{-a}(2k)!\sum^{2k}_{n=0}\frac{a^n}{n!}$$ Hence $$I=\sum^{\infty}_{k=0}\frac{(-1)^k}{(2k+1)}\left(\sum^{2k}_{n=0}\frac{a^n}{n!}+\sum^{2k}_{n=0}\frac{(-a)^n}{n!}\right)$$ Or equivalently $$I(a)=2\sum^{\infty}_{k=0}\frac{(-1)^k}{2k+1}\sum^{k}_{n=0}\frac{a^{2n}}{(2n)!}.$$