Evaluate integral $\int_t^\infty (1-3x^{-4})e^{-x^2/2}dx$ I am trying to evaluate the integral
$$\int_t^\infty (1-3x^{-4})e^{-x^2/2}\,\mathrm{d}x.$$
This is as far as I could get before getting stuck.
$$
\begin{align}
\int_t^\infty (1-3x^{-4})e^{-x^2/2}\,\mathrm{d}x&=\int_t^\infty e^{-x^2/2}\,\mathrm{d}\left(x+\frac{1}{x^3}\right) \\
&=-\left(t+\frac{1}{t^3}\right)e^{-t^2/2}-\int_t^\infty \left(x+\frac{1}{x^3}\right)\,\mathrm{d}(e^{-x^2/2}) \\
&=-\left(t+\frac{1}{t^3}\right)e^{-t^2/2}+\int_t^\infty \left(x^2+\frac{1}{x^2}\right)e^{-x^2/2}\,\mathrm{d}x \\
\int_t^\infty \left(x^2+\frac{1}{x^2}\right)e^{-x^2/2}\,\mathrm{d}x&=\int_t^\infty e^{-x^2/2}\,\mathrm{d}\left(\frac{1}{3}x^3-\frac{1}{x}\right)
\end{align}
$$

$$I=I_1-I_2$$
$$I_1=\int_{t}^{\infty} e^{-x^2/2} dx= \sqrt{2} \int_{t/\sqrt{2}}^{\infty} e^{-z^2} dz=\sqrt{\frac{\pi}{2}}~Erfc(t/\sqrt{2})$$
$$I_2=3\int_{t}^{\infty} x^{-4}~ e^{-x^2/2} dx$$
Let $x^2=2u, xdx=du$, then
$$I_2=\frac{3}{4\sqrt{2}} \int_{t^2/2}^{\infty} u^{-5/2} e^{-u} du$$
Next, twice integration by parts leads to
$$I_2=\frac{3}{4\sqrt{2}}\left[-\frac{2}{3} e^{-u}~ u^{-3/2}+\frac{4}{3} e^{-u} u^{-1/2}+\frac{4}{3}\sqrt{\pi} ~Erf(\sqrt{u})~du \right]_{t^2/2}^{\infty}$$
Usw $Erfc(z)=1-Erf(z),~ Erf(\infty)=1$
$$I=\left(\frac{1}{t}-\frac{1}{t^3} \right) e^{-t^2/2}.$$
