How to integrate $\int_{1.96}^{\infty}e^{-\frac {x^2} {2}}\,dx$ I can't integrate it using the methods I know. 
 A: If you remember the z-scores of a standard normal distribution, $\int_{1.96}^\infty\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}=1-N(1.96)\approx0.025$, where $N(.)$ is the CDF of a standard normal distribution.
So the answer is approximately $0.025\times \sqrt{2\pi}\approx0.025*2.5=0.0625$. 
A: Concerning the antiderivative first $$\int e^{-c x^2}\,dx=\frac{\sqrt{\pi } }{2 \sqrt{c}}~~\text{erf}\left(\sqrt{c} x\right)$$ where appears the error function. Now, for the integral $$\int_b^\infty e^{-c x^2}\,dx=\frac{\sqrt{\pi } }{2 \sqrt{c}}\Big(1-\text{erf}\left(b \sqrt{c}\right)\Big)$$ If you go to the Wikipedia page, you will find a series of more or less accurate numerical approximations of the error function. One which is quite simple is $$\text{erf(x)}\approx \text{sgn}(x)\sqrt{1-e^{-\frac{x^2 \left(a x^2+\frac{4}{\pi }\right)}{a x^2+1}}}$$ where $$a=\frac{8 (\pi -3)}{3 (4-\pi ) \pi }\approx 0.140012$$ Applied to your case $c=\frac12$, $b=1.96$, the above formula gives $\approx 0.06225$ while the exact value would be $\approx 0.06266$ which is not too bad for practical purposes.
Using $a=0.148009$ would give a better approximation ($\approx 0.06257$).
A: I have an idea in solving it but not sure if it is useful or no!
$\begin{array}{l}
u = \frac{{{x^2}}}{2} \Rightarrow S = \int\limits_{1.96}^\infty  {{e^{ - u}}\frac{1}{{\sqrt u }}} du\\
t = u - 1.96 \Rightarrow S = \int\limits_0^\infty  {{e^{ - t}}\frac{{{e^{ - 1.96}}}}{{\sqrt {t + 1.96} }}} dt
\end{array}$
Then use Laplace transform definition to find the integral.
