How to solve gaussian integral $\int_{-\infty} ^ \infty \sin(x) e^{-m(x-a)^2} dx $ $$\int_{-\infty} ^ \infty \sin(x) e^{-m(x-a)^2} dx$$
where $a$ and $m$ are constants. Is there an alternate way of solving this without using the Euler's formula or without breaking down sin into exponential form.
 A: $I=\int_{-\infty} ^ \infty \sin(x) e^{-m(x-a)^2} dx = \int_{-\infty} ^ \infty \sin(X+a) e^{-m X^2} dX \quad $ with change of variable $x=X+a$
$I=\sin(a)\int_{-\infty} ^ \infty \cos(X) e^{-m X^2} dX +\cos(a)\int_{-\infty} ^ \infty \sin(X) e^{-m X^2} dX$ 
$\int_{-\infty} ^ \infty \sin(X) e^{-m X^2} dX=0$
$\int_{-\infty} ^ \infty \cos(X) e^{-m X^2} dX = \sqrt{\frac{\pi}{m}}e^{-\frac{1}{4m}}$ 
$$\int_{-\infty} ^ \infty \sin(x) e^{-m(x-a)^2} dx =\sin(a)\sqrt{\frac{\pi}{m}}e^{-\frac{1}{4m}}$$
A: I will postpone the question of convergence for now. Your (real-valued) integral is 
$$
S = \int_{-\infty}^{\infty} \sin(x) e^{-m(x-a)^2} \, \mathrm{d} x \, .
$$
I assume that $m>0$. It can also be written using a complex exponential as
$$
S = \Im \left [ \int_{-\infty}^{\infty} e^ { ix - m(x-a)^2 } \, \mathrm{d} x   \right ] = \Im \left [ I \right ] \, ,
$$
where $I$ is the newly defined complex-values integral. Completing the square, I get
$$
I = \int_{-\infty}^{\infty} e^{ -m \left [ x - \left ( a + \frac{i}{2m} \right ) \right ]^2 + \left ( ia - \frac{1}{4m} \right ) } \, \mathrm{d} x \, .
$$
Changing variables to $y = x - \left ( a + \frac{i}{2m} \right )$,
$$
I = \int_{-\infty}^{\infty} e^{ -m y^2 + \left ( ia - \frac{1}{4m} \right ) } \, \mathrm{d} y  = \sqrt{\frac{\pi}{m}} e^{ ia - \frac{1}{4m}} \, .
$$
Looking at the real and imaginary part separately, I get
$$
I = C + iS = \sqrt{\frac{\pi}{m}} e^{-\frac{1}{4m}} \cos(a) + i \sqrt{\frac{\pi}{m}} e^{-\frac{1}{4m}} \sin(a) \, .
$$
Result:
Therefore it follows that
$$
S = \int_{-\infty}^{\infty} \sin(x) e^ { - m(x-a)^2 } \, \mathrm{d} x = \sqrt{\frac{\pi}{m}} e^{-\frac{1}{4m}} \sin(a) \, ,
$$
and you get for free
$$
C = \int_{-\infty}^{\infty} \cos(x) e^ {- m(x-a)^2 } \, \mathrm{d} x = \sqrt{\frac{\pi}{m}} e^{-\frac{1}{4m}} \cos(a) \, .
$$
