This Quantum World/Appendix/Indefinite integral

The indefinite integral
How do we add up infinitely many infinitesimal areas? This is elementary if we know a function $$F(x)$$ of which $$f(x)$$ is the first derivative. If $$f(x)=\frac{dF}{dx}$$ then $$dF(x)=f(x)\,dx$$ and



\int_a^b f(x)\,dx=\int_a^b dF(x)=F(b)-F(a).$$

All we have to do is to add up the infinitesimal amounts $$dF$$ by which $$F(x)$$ increases as $$x$$ increases from $$a$$ to $$b,$$ and this is simply the difference between $$F(b)$$ and $$F(a).$$

A function $$F(x)$$ of which $$f(x)$$ is the first derivative is called an integral or antiderivative of $$f(x).$$ Because the integral of $$f(x)$$ is determined only up to a constant, it is also known as indefinite integral of $$f(x).$$ Note that wherever $$f(x)$$ is negative, the area between its graph and the $$x$$ axis counts as negative.

How do we calculate the integral $$I=\int_a^b dx\,f(x)$$ if we don't know any antiderivative of the integrand $$f(x)$$? Generally we look up a table of integrals. Doing it ourselves calls for a significant amount of skill. As an illustration, let us do the Gaussian integral



I=\int_{-\infty}^{+\infty}dx\,e^{-x^2/2}.$$

For this integral someone has discovered the following trick. (The trouble is that different integrals generally require different tricks.) Start with the square of $$I$$:



I^2=\int_{-\infty}^{+\infty}\!dx\,e^{-x^2/2}\int_{-\infty}^{+\infty}\!dy \,e^{-y^2/2}= \int_{-\infty}^{+\infty}\!\int_{-\infty}^{+\infty}\!dx\,dy\,e^{-(x^2+y^2)/2}. $$

This is an integral over the $$x{-}y$$ plane. Instead of dividing this plane into infinitesimal rectangles $$dx\,dy,$$ we may divide it into concentric rings of radius $$r$$ and infinitesimal width $$dr.$$ Since the area of such a ring is $$2\pi r\,dr,$$ we have that



I^2=2\pi\int_0^{+\infty}\!dr\,r\,e^{-r^2/2}.$$

Now there is only one integration to be done. Next we make use of the fact that $$\frac{d\,r^2}{dr}=2r,$$ hence $$dr\,r=d(r^2/2),$$ and we introduce the variable $$w=r^2/2$$:



I^2=2\pi\int_0^{+\infty}\!d\left({r^2/2}\right)e^{-r^2/2}= 2\pi\int_0^{+\infty}\!dw\,e^{-w}.$$

Since we know that the antiderivative of $$e^{-w}$$ is $$-e^{-w},$$ we also know that



\int_0^{+\infty}\!dw\,e^{-w}=(-e^{-\infty})-(-e^{-0})=0+1=1.$$

Therefore $$I^2=2\pi$$ and



\int_{-\infty}^{+\infty}\!dx\,e^{-x^2/2}=\sqrt{2\pi}.$$

Believe it or not, a significant fraction of the literature in theoretical physics concerns variations and elaborations of this basic Gaussian integral.

One variation is obtained by substituting $$\sqrt{a}\,x$$ for $$x$$:



\int_{-\infty}^{+\infty}\!dx\,e^{-ax^2/2}=\sqrt{2\pi/a}.$$

Another variation is obtained by thinking of both sides of this equation as functions of $$a$$ and differentiating them with respect to $$a.$$ The result is



\int_{-\infty}^{+\infty}dx\,e^{-ax^2/2}x^2=\sqrt{2\pi/a^3}.$$