Integration by parts. Which is the simplest way for this integral? Which is the easiest way to compute the following integral by parts?
$$\displaystyle\int\limits_{x}^{\infty}\dfrac{2}{\sqrt{\pi}}e^{-s^2}ds$$?
Given that $\int f'(x)\cdot g(x)=f(x)\cdot g(x)-\int f'(x)\cdot g(x) +c$
could I choose $f(x)=e^{-s^2}$ and $g'(x)=1$?
Which is the simplest way to the solution of the integral by parts?
 A: To nitpick a few comments, it is impossible to find the antiderivative of this function in terms of elementary functions. However, it is perfectly possible to compute this integral. 
First, recall that the integral is a linear operator, which is a fancy term for saying that $\int af(x)dx = a \int f(x)dx$ and $\int (f(x)+g(x))dx =  \int f(x)dx+\int g(x)dx$. 
So to start, we can express the integral as:
$$\frac{2}{\sqrt{\pi}}\int_x^\infty e^{-s^2}ds$$
If you wanted to find the antiderivative of this function, you would need a series expansion, and to express the result in terms of a series. However, it is perfectly possible to find this specific integral with a little trick.
Let's set $I = \int_x^\infty e^{-s^2}ds$ (we'll multiply by the constant fraction in the end). Let's also restrict this answer to $x \geq 0$: it's easy to observe that $e^{-x^2}$ is even, so all we would need to do to find the answer for $x<0$ is compute $2*\int_0^\infty e^{-s^2}ds-\int_{|x|}^\infty e^{-s^2}ds$. Observe that $I = \int_x^\infty e^{-t^2}dt$ as well, so:
$$I^2 = \int_x^\infty e^{-s^2}ds*\int_x^\infty e^{-t^2}dt$$
$t$ and $s$ are constant realative to each other, so:
$$I^2 = \int_x^\infty \int_x^\infty e^{-s^2}e^{-t^2}dsdt = \int_x^\infty \int_x^\infty e^{-(s^2+t^2)}dsdt$$
Observe that, if we were to plot the region of integration with respect to s and t, we would find that we are integrating from the quarter circle with radius $x$ in quadrant I to the quarter circle with radius $c$ in quadrant I, where $c$ increases without bound. 
So how does this help us? Well, we can integrate much more easily if we use polar coordinates, because while the differential in linear coordinates is $ds*dt$ (the area of a small rectangle in those coordinates), if we integrate with variables $r = \sqrt{s^2+t^2}$ and $\theta  = arctan(t/s)$, the differential is $r*dr*ds$. So:
$$I^2 = \int_x^\infty \int_0^{\pi/2} e^{-r^2}*r*d\theta dr = \frac{\pi}{2}\int_x^\infty e^{-r^2}*rdr = -\frac{\pi}{4}[e^{-r^2}]_x^\infty = \frac{\pi}{4} e^{-x^2}$$
So, for $x \geq 0$
$$I = \frac{\sqrt{\pi}}{2} e^{-\frac{x^2}{2}}$$
Which means that for $x \geq 0$:
$$\frac{2}{\sqrt{\pi}}\int_x^\infty e^{-s^2}ds = e^{-\frac{x^2}{2}}$$
And for $x < 0$:
$$\frac{2}{\sqrt{\pi}}\int_x^\infty e^{-s^2}ds = 2*e^{-\frac{0^2}{2}}-e^{-\frac{x^2}{2}} = 2-e^{-\frac{x^2}{2}}$$
Note that this does not contradict the idea that $e^{-x^2}$ does not have an elementary antiderivative. You cannot differentiate the result here to get back the original function. This can only be used to calculate actual numerical areas given that one of your upper bound is infinite.
A: You won't be able to use integration by parts on this; the function doesn't have an antiderivative which can be expressed with elementary ("normal") functions.
You may find it interesting to note that at $x=0$, your integral evaluates to $1$; this is known as the Gaussian integral.
