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I have this improper integral:

$$I = \int_{-\infty}^{\infty} x^4 e^{\frac{-x^2}{2t}} \ dx$$

I used a substitution to get ($u=\frac{x^2}{2t}$):

$$I = \int_{-\infty}^{\infty} u^{\frac{3}{2}} e^{-u} \ du$$

(note I have taken the t's and the 2's to the front and left them out for now as they are constants)

However I am having trouble solving it. Ideally I was thinking of using the gamma integral however the limits are not correct. I need them to be between $0$ and $\infty$, I was thinking of doing this by trying to make the function even so I can change the limits. I think this should be right as this would yield $2*\Gamma(\frac{5}{2})$ which is the answer I wanted. However the function above is not even to my knowledge so I am stuck!

Could anyone help me out? I don't need an answer just leave a comment please!

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    $\begingroup$ Hint: Use the evenness of $x^4e^{-x^2/(2t)}$ to write $$\int_{-\infty}^\infty x^4e^{-x^2/(2t)}\,dx=2\int_0^\infty x^4e^{-x^2/(2t)}\,dx$$ before making the substitution $u=x^2/(2t)$. (The integral you derived is incorrect: $u^{3/2}$ is either undefined or imaginary for $u\lt0$, and the $e^{-u}$ would make the integral over the negative reals diverge anyway. On a related note, the original integral only makes sense if $t\gt0$; it diverges if $t\lt0$.) $\endgroup$ Jan 15 at 11:50
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The function is even, hence you can write it as

$$2\int_0^{+\infty} x^4 e^{-x^2/2t}\ \text{d}t$$

Using your substitution (please be careful of what you get), you can deal it via Gamma function, obtaining the result $$3 \sqrt{2 \pi } t^{5/2}$$

Details on the substitution

$$z = \dfrac{x^2}{2t} ~~~~~~~ x = \sqrt{2tz} ~~~~~~~ \text{d}x = \dfrac{t^{1/2}}{\sqrt{2z}}\ \text{d}z$$

Hence the integral becomes

$$2\int_0^{+\infty} 4 t^2z^2 e^{-z}\left(\dfrac{t^{1/2}}{\sqrt{2z}}\right)\ \text{d}z$$

$$\dfrac{8}{\sqrt{2}}t^{5/2}\int_0^{+\infty} z^{3/2}e^{-z}\ \text{d}z$$

Whence the above result.

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$$I=\int x^4 e^{-x^2/2t}\ \text{d}t=\\Integrate -by- parts{\int}\mathtt{f}\mathtt{g}' = \mathtt{f}\mathtt{g} - {\int}\mathtt{f}'\mathtt{g}\\=-tx^3\mathrm{e}^{-\frac{x^2}{2t}}-{\displaystyle\int}-3tx^2\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x$$Now solving ${\displaystyle\int}-3tx^2\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x==-\class{steps-node}{\cssId{steps-node-1}{3t}}{\cdot}{\displaystyle\int}x^2\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x$ $$=-tx\mathrm{e}^{-\frac{x^2}{2t}}-{\displaystyle\int}-t\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x$$ Now solving ${\displaystyle\int}-t\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x$
Substitute $u=\dfrac{x}{\sqrt{2}\sqrt{t}}$
$${\displaystyle\int}-t\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x=-\class{steps-node}{\cssId{steps-node-2}{\dfrac{\sqrt{{\pi}}t^\frac{3}{2}}{\sqrt{2}}}}{\displaystyle\int}\dfrac{2\mathrm{e}^{-u^2}}{\sqrt{{\pi}}}\,\mathrm{d}u$$ note that ${\displaystyle\int}\dfrac{2\mathrm{e}^{-u^2}}{\sqrt{{\pi}}}\,\mathrm{d}u$ This is a special integral ,Gauss error function which called $=erf(u)$so $$-\class{steps-node}{\cssId{steps-node-3}{\dfrac{\sqrt{{\pi}}t^\frac{3}{2}}{\sqrt{2}}}}{\displaystyle\int}\dfrac{2\mathrm{e}^{-u^2}}{\sqrt{{\pi}}}\,\mathrm{d}u==-\dfrac{\sqrt{{\pi}}t^\frac{3}{2}\operatorname{erf}\left(\frac{x}{\sqrt{2}\sqrt{t}}\right)}{\sqrt{2}}$$so plug in the second step $$-tx^3\mathrm{e}^{-\frac{x^2}{2t}}-{\displaystyle\int}-3tx^2\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x=\\=\dfrac{3\sqrt{{\pi}}t^\frac{5}{2}\operatorname{erf}\left(\frac{x}{\sqrt{2}\sqrt{t}}\right)}{\sqrt{2}}-tx^3\mathrm{e}^{-\frac{x^2}{2t}}-3t^2x\mathrm{e}^{-\frac{x^2}{2t}}$$ an ,finally $${\displaystyle\int}x^4\mathrm{e}^{-\frac{x^2}{2t}}\,\mathrm{d}x=\\=\dfrac{3\sqrt{{\pi}}t^\frac{5}{2}\operatorname{erf}\left(\frac{x}{\sqrt{2}\sqrt{t}}\right)}{\sqrt{2}}-tx^3\mathrm{e}^{-\frac{x^2}{2t}}-3t^2x\mathrm{e}^{-\frac{x^2}{2t}}+C$$ now apply limit bounds to find numerical value $$I=I = \int_{-\infty}^{\infty} x^4 e^{\frac{-x^2}{2t}} \ dx=\dfrac{3\sqrt{{\pi}}t^\frac{5}{2}}{\sqrt{2}}$$

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$$ I = \int_{-\infty}^\infty u^{3/2}e^{-u}du \\ \implies I = \int_0^\infty u^{3/2}e^{-u}du + \int_{-\infty}^0 u^{3/2}e^{-u}du $$ Changing the sign of the limits on the second term and swapping them has no effect, as both introduce one minus sign, so we can write $$ I = 2\int_0^\infty u^{3/2}e^{-u}du = 2\Gamma(5/2). $$

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Your substitution is invalid as it stands because $x^2/2t$ is not bijective on the integration domain. The integral is, however, twice the integral from $0$ to $\infty$, whereby the substitution and gamma definition can be applied and the result is $(2t)^{5/2}\cdot\frac{3\sqrt\pi}4$.

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