<s>[INST] Compute $I = \int_{0}^{\infty} \frac{8}{\gamma^{4}} x^{3} e^{-\frac{2}{\gamma^{2}}x^{2}} e^{jxt} dx$ I want to evaluate the following integral:
$$I = \int_{0}^{\infty} \frac{8}{\gamma^{4}} x^{3} e^{-\frac{2}{\gamma^{2}}x^{2}} e^{jxt} dx $$
Where $\gamma \in \mathbb{R}$ and $j = \sqrt{-1}$. 

The first thing I do is let:
$$ u = \frac{2}{\gamma} x  \quad \Rightarrow  \quad  du = \frac{2}{\gamma} dx$$
$$ x = \frac{\gamma}{2} u  \quad \Rightarrow  \quad  dx = \frac{\gamma}{2} du$$
After substituting and cancellation, we get:
$$ I = \frac{1}{2} \int_{0}^{\infty} u^{3} e^{-\frac{1}{2} u^{2} + j \frac{1}{2} \gamma tu} du$$
We complete the square, giving:
$$ I = \frac{1}{2} \int_{0}^{\infty} u^{3} e^{-\frac{1}{2} (u - j\frac{1}{2} \gamma t)^{2} - \frac{1}{8} \gamma^{2} t^{2}} du$$
$$ I = \frac{1}{2} e^{-\frac{1}{8} \gamma^{2} t^{2}} \int_{0}^{\infty} u^{3} e^{-\frac{1}{2} (u - j\frac{1}{2} \gamma t)^{2}} du$$
This is where I am unsure of what to do. My approach is as follows:

We want to solve:
$$ I = \frac{1}{2}e^{-\frac{1}{8} \gamma^{2} t^{2}} \underset{R \rightarrow \infty}{lim}  \int_{0}^{R} u^{3} e^{-\frac{1}{2} (u - j\frac{1}{2} \gamma t)^{2}} du$$
We let:
$$ z = u - j \frac{1}{2} \gamma t  \quad \Rightarrow  \quad  dy = du$$
$$ u = y + j\frac{1}{2} \gamma t  \quad \Rightarrow  \quad  du = dy$$
$$ b = R - j\frac{1}{2} \gamma t \quad a = -j \frac{1}{2} \gamma t $$
Therefore:
$$ I = \frac{1}{2}e^{-\frac{1}{8} \gamma^{2} t^{2}} \underset{R \rightarrow \infty}{lim}  \int_{ -j \frac{1}{2} \gamma t}^{R - j \frac{1}{2} \gamma t} (y- j \frac{1}{2} \gamma t)^{3} e^{-\frac{1}{2} y^{2}} dy$$
We create a contour $\Gamma = C_{1} + C_{2} + C_{3} + C_{4} = [-j \frac{1}{2} \gamma t, R-j \frac{1}{2} \gamma t] + [R-j \frac{1}{2} \gamma t, R] + [R, 0] + [0, -j \frac{1}{2} \gamma t]$
Since $ \underset{y \rightarrow \infty}{lim}\frac{(y - j \frac{1}{2} \gamma t)^{3}}{e^{\frac{1}{2}y^{2}}} = 0  $, the integral over the vertical contour $C_{3}$ is zero. 
However I still need to evaluate the following integrals:
$$ I_{1} = \underset{R \rightarrow \infty}{lim}  \int_{0}^{R} (y- j \frac{1}{2} \gamma t)^{3} e^{-\frac{1}{2} y^{2}} dy $$
$$ I_{2} = \int_{-j \frac{1}{2} \gamma t}^{0} (y- j \frac{1}{2} \gamma t)^{3} e^{-\frac{1}{2} y^{2}} dy $$
This is where I am getting stuck.

Is this the right approach? Am I on the right track? How can I solve those two integrals? How can I solve my original problem?
Thanks!
 [/INST] Hint: find $$\int_0^\infty e^{-2 x^2/\gamma^2} e^{jxt}\; dx $$
and differentiate $3$ times with respect to $t$.
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