Calculate $\int_{-T}^T\sin(x-a)\cdot\sin(x-b)~e^{-k~(x-a)(x-b)}~dx\quad$ Calculate
$$\displaystyle\int_{-T}^T\sin(x-a)\cdot\sin(x-b)~e^{-k~(x-a)(x-b)}~dx\quad$$
I have no idea how to proceed. Any suggestions please? Here $T>0$.
 A: Here is the path to completion - not the answer. I will be assuming that $a,b,k$ - are real numbers.
First, observe that $$\sin(x-a)\sin(x-b) = \frac{1}{2}(\cos(b-a)-\cos(2x-(a+b)))$$
Now, split integral in 3 parts:$$I_0 = \frac{\cos(a-b)}{2}\int_{-T}^{T}e^{-k((x-\frac{a+b}{2})^2 - \frac{a^2+b^2}{4})}dx$$ 
Other two integrals come from presentation of $\cos$ as a sum of two exponents: $$I_1 = -\frac{1}{4}\int_{-T}^{T}e^{-k((x-\frac{a+b}{2})^2 - \frac{a^2+b^2}{4}) + 2i(x - \frac{a+b}{2})}dx$$
$$I_2 = -\frac{1}{4}\int_{-T}^{T}e^{-k((x-\frac{a+b}{2})^2 - \frac{a^2+b^2}{4}) - 2i(x - \frac{a+b}{2})}dx$$
By variables substitution, $I_0$ can be presented as combination of error functions - $$erf(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt$$ if $k >0$ or Fresnel integrals if $k < 0$: $$C(x) = \int_{0}^{x}\cos(t^2)dt; S(x) = \int_{0}^{x}\sin(t^2)dt$$
In $I_1$ and $I_2$ substitute variables $u = x-\frac{a+b}{2}$ $$-ku^2 \pm 2iu = -(u\sqrt{k}\mp\frac{i}{\sqrt{k}})^2 - \frac{1}{k}$$ 
Let $y = u\sqrt{k}\mp\frac{i}{\sqrt{k}}$ 
Now we will find that integration line shifted to complex plane. 
All functions are analytic without poles on our contour and we can move line of the contour in a way to present integral as a sum of imaginary error functions. 
Combining altogether gives us linear combination of error functions (or Fresnel functions, depends on k) and imaginary error functions. 
Upon completion we shall pay attention to select right signs of imaginary parts to let them compensate as initial integral is real.
I would not dare to complete all above as it require lots of tedious computations. 
Hope it help.
