Convolution Integral problem I'm having a hard time to do this exercise. I'm using the definition for convolution but I'm stuck at the integral. Thanks in advance.

For $t>0$ consider $f_a(x)=\frac{1}{\sqrt{4 \pi a}}e^{-\frac{x^2}{4a}}$.
Show that $f_a*f_b=f_{a+b}$

By definition: $(f*g)(x)$= $\int_{-\infty}^{+\infty} f(u)g(x-u) du$
One computes,
$(f_a*f_b)(x)=$ $\int_{-\infty}^{+\infty} \frac{1}{\sqrt{4 \pi a}}e^{-\frac{u^2}{4a}} \frac{1}{\sqrt{4 \pi b}}e^{-\frac{(x-u)^2}{4b}} du$
$=\frac{1}{\sqrt{4 \pi a}}\frac{1}{\sqrt{4 \pi b}} \int_{-\infty}^{+\infty} e^{-\frac{u^2}{4a}}e^{-\frac{(x-u)^2}{4b}}$
I can't solve this integral. I tried using Gaussian Integral but still cant solve it.
 A: One of the most useful elementary tricks to know when dealing with Gaussian-type integrals is completing the square.
On that note, we calculate
\begin{align}
f_a*f_b(x)
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{y^2}{4a}-\frac{(x-y)^2}{4b}\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{1}{4a}\left(y^2+\frac{a}{b}(x-y)^2\right)\right)\,\mathrm{d}y\\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{1}{4a}\left(\left(1+\frac{a}{b}\right)y^2-\frac{2a}{b}xy+\frac{a}{b}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}\left(y^2-\frac{2a}{a+b}xy+\frac{a}{a+b}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}\left(\left(y-\frac{a}{a+b}x\right)^2+\frac{ab}{(a+b)^2}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}\left(y^2+\frac{ab}{(a+b)^2}x^2\right)\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\exp\left(-\frac{x^2}{4(a+b)}\right)
\int_{-\infty}^\infty\exp\left(
-\frac{a+b}{4ab}y^2\right)\,\mathrm{d}y \\
& =\frac{1}{4\pi \sqrt{ab}}\sqrt{\frac{4ab}{a+b}}\exp\left(-\frac{x^2}{4(a+b)}\right)
\int_{-\infty}^\infty e^{-z^2}\,\mathrm{d}z \\
&= \frac{1}{\sqrt{4\pi(a+b)}}\exp\left(-\frac{x^2}{4(a+b)}\right) \\
&=f_{a+b}(x).
\end{align}
Of course, using the Fourier transform, we can simplify this calculation somewhat – if you haven't read about the Fourier transform (and its action on Gaussians), I highly recommend you take a look. The basic techniques used in calculating the Fourier transform of a Gaussian are in fact very closely analogous to the calculations above.
A: https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables#Proof_using_convolutions
You can find the proof here using convolutions - just use $\sigma^2_X = 2a$ and $\sigma^2_Y = 2b$, $\mu_X = 0$ and $\mu_Y = 0$. It involves making a substitution involving $a$ and $b$, then completing the square.
