# 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 ${\displaystyle F(x)}$ of which ${\displaystyle f(x)}$ is the first derivative. If ${\displaystyle f(x)={\frac {dF}{dx}}}$ then ${\displaystyle dF(x)=f(x)\,dx}$ and

${\displaystyle \int _{a}^{b}f(x)\,dx=\int _{a}^{b}dF(x)=F(b)-F(a).}$

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

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

How do we calculate the integral ${\displaystyle I=\int _{a}^{b}dx\,f(x)}$ if we don't know any antiderivative of the integrand ${\displaystyle f(x)}$? 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

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

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

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

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

${\displaystyle I^{2}=2\pi \int _{0}^{+\infty }\!dr\,r\,e^{-r^{2}/2}.}$

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

${\displaystyle I^{2}=2\pi \int _{0}^{+\infty }\!d\left({r^{2}/2}\right)e^{-r^{2}/2}=2\pi \int _{0}^{+\infty }\!dw\,e^{-w}.}$

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

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

Therefore ${\displaystyle I^{2}=2\pi }$ and

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

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 ${\displaystyle {\sqrt {a}}\,x}$ for ${\displaystyle x}$:

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

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

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