How to solve gaussian integral $\int_{-\infty} ^ \infty \sin(x) e^{-m(x-a)^2} dx $

Question

$$\int_{-\infty} ^ \infty \sin(x) e^{-m(x-a)^2} dx$$

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.

Answer 1

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 $$ C = \int_{-\infty}^{\infty} \cos(x) e^ {- m(x-a)^2 } \, \mathrm{d} x = \sqrt{\frac{\pi}{m}} e^{-\frac{1}{4m}} \cos(a) \, . $$

Answer 2

$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 $x=X+a$

$I=\sin(a)\int_{-\infty} ^ \infty \cos(X) e^{-m X^2} dX +\cos(a)\int_{-\infty} ^ \infty \sin(X) e^{-m X^2} dX$

$\int_{-\infty} ^ \infty \sin(X) e^{-m X^2} dX=0$

$\int_{-\infty} ^ \infty \cos(X) e^{-m X^2} dX = \sqrt{\frac{\pi}{m}}e^{-\frac{1}{4m}}$

$$\int_{-\infty} ^ \infty \sin(x) e^{-m(x-a)^2} dx =\sin(a)\sqrt{\frac{\pi}{m}}e^{-\frac{1}{4m}}$$