What is the antiderivative of $\quad$$t^2e^{-\frac{1}{2}t^2}$ ?
$\displaystyle\int t^2e^{-\frac{1}{2}t^2}\,dt=\displaystyle\int{t}_{}te^{-\frac{1}{2}t^2}\,dt=-te^{-\frac{1}{2}t^2}\Big|_{?}^?+\int e^{-\frac{1}{2}t^2}$
Is there no explicit antiderivative, Thanks in advance.
On
Using IBP by letting $u=t$, $du=dt$, and $$ dv=te^{-\large\frac12t^2}\ dt\quad\Rightarrow\quad v=\int te^{-\large\frac12t^2}\ dt=-e^{-\large\frac12t^2}, $$ yield $$ \int t^2e^{-\large\frac12t^2}\ dt=-te^{-\large\frac12t^2}+\int e^{-\large\frac12t^2}\ dt=-te^{-\large\frac12t^2}+\sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{x}{\sqrt{2}}\right). $$ The last part is the error function (also called the Gauss error function). Assuming the integral in the middle as $$ \int_0^x e^{-\large\frac12t^2}\ dt=\sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{x}{\sqrt{2}}\right). $$
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What you wrote, integrating by parts, is kind of correct. The integral of the Gaussian is impossible to express in terms of simple functions, but is well known, it is called error function.
It has the defining property: $$ \mathrm{erf}\bigg(\frac{x}{\sqrt{2}}\bigg) =\sqrt{\frac{2}{\pi}} \int_0^x e^{-\frac{1}{2}t^2}\,dt. $$
Therefore the antiderivative is, for the correct calculation you did: $$ -t\,e^{-\frac{1}{2}t^2} + \sqrt{\frac{\pi}{2}} \,\mathrm{erf}\bigg( \frac{x}{\sqrt{2}} \bigg) + c. $$
There is no elementary anti-derivative for $e^{-t^2/2}$, but there are several special functions such as the error function $$ \int e^{-t^2/2}\,\mathrm{d}t=\sqrt{\frac\pi2}\,\mathrm{erf}\left(\frac t{\sqrt2}\right)+C $$ Note that $\mathrm{erf}(0)=0$. Your integration by parts is correct.
The error function above is defined as $$ \mathrm{erf}(x)=\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm{d}t $$ Thus, $\mathrm{erf}(x)$ is odd and $\lim\limits_{x\to\infty}\mathrm{erf}(x)=1$.