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!

Well, in general we have:
$$\mathcal{I}_\text{n}:=\int_0^{\infty}x^\text{n}\cdot\exp\left(\alpha \cdot x^{\text{n}-1}\right)\cdot\exp\left(\beta\cdot x\right)\space\text{d}x\tag1$$
Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:
$$\mathcal{I}_\text{n}=\int_0^\infty\mathcal{L}_x\left[\exp\left(\alpha \cdot x^{\text{n}-1}\right)\cdot\exp\left(\beta\cdot x\right)\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[x^\text{n}\right]_{\left(\text{s}\right)}\space\text{ds}\tag2$$
Using the properties of the Laplace transform:


*

*$$\mathcal{L}_x\left[\exp\left(\alpha \cdot x^{\text{n}-1}\right)\cdot\exp\left(\beta\cdot x\right)\right]_{\left(\text{s}\right)}=\mathcal{L}_x\left[\exp\left(\alpha \cdot x^{\text{n}-1}\right)\right]_{\left(\text{s}-\beta\right)}\tag3$$

*$$\mathcal{L}_x\left[\exp\left(\alpha \cdot x^{\text{n}-1}\right)\right]_{\left(\text{s}-\beta\right)}=\sum_{\text{k}=0}^\infty\frac{\alpha^\text{k}}{\text{k}!}\cdot\mathcal{L}_x\left[x^{\text{k}\left(\text{n}-1\right)}\right]_{\left(\text{s}-\beta\right)}=\sum_{\text{k}=0}^\infty\frac{\alpha^\text{k}}{\text{k}!}\cdot\frac{\Gamma\left(1+\text{k}\left(\text{n}-1\right)\right)}{\left(\text{s}-\beta\right)^{1+\text{k}\left(\text{n}-1\right)}}\tag4$$

*$$\mathcal{L}_x^{-1}\left[x^\text{n}\right]_{\left(\text{s}\right)}=\frac{1}{\Gamma\left(-\text{n}\right)}\cdot\frac{1}{\text{s}^{1+\text{n}}}\tag5$$
So, we get:
$$\mathcal{I}_\text{n}=\int_0^\infty\left\{\sum_{\text{k}=0}^\infty\frac{\alpha^\text{k}}{\text{k}!}\cdot\frac{\Gamma\left(1+\text{k}\left(\text{n}-1\right)\right)}{\left(\text{s}-\beta\right)^{1+\text{k}\left(\text{n}-1\right)}}\right\}\cdot\frac{1}{\Gamma\left(-\text{n}\right)}\cdot\frac{1}{\text{s}^{1+\text{n}}}\space\text{ds}=$$
$$\sum_{\text{k}=0}^\infty\frac{\alpha^\text{k}}{\text{k}!}\cdot\frac{\Gamma\left(1+\text{k}\left(\text{n}-1\right)\right)}{\Gamma\left(-\text{n}\right)}\cdot\int_0^\infty\frac{\text{s}^{-1-\text{n}}}{\left(\text{s}-\beta\right)^{1+\text{k}\left(\text{n}-1\right)}}\space\text{ds}\tag6$$
Using the property:
$$\int_0^\infty\frac{x^\text{m}}{\left(x-t\right)^\text{p}}\space\text{d}x=\frac{1}{\left(-t\right)^{\text{p}-1-\text{m}}}\cdot\frac{\Gamma\left(1+\text{m}\right)\cdot\Gamma\left(\text{p}-\text{m}-1\right)}{\Gamma\left(\text{p}\right)}\tag7$$
Gives:
$$\mathcal{I}_\text{n}=\sum_{\text{k}=0}^\infty\frac{\alpha^\text{k}}{\text{k}!}\cdot\left(-1\right)^{\text{k}-\text{n}\left(1+\text{k}\right)-1}\cdot\beta^{\text{k}-\text{n}\left(1+\text{k}\right)-1}\cdot\Gamma\left(1+\text{k}\left(\text{n}-1\right)+\text{n}\right)\tag8$$
