Use an appropriate series expansion

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Use an appropriate series expansion

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[QUICK DESCRIPTION]

Suppose we have a general integral of the form
[math] \int_X f(x) d\mu(x) [/math]
and we know how to expand $f$ in a series
[math]f(x) \sim \sum c_k h_k(x),[/math]
where each one of the base functions $h_k$ is [[Which integrals are simpler to integrate|easy to integrate]]. Then [[interchange integrals or sums|interchanging the integral and the sum]] we can simplify the original integral to something like

[math] \int_X f(x) d\mu(x) = \sum _k c_k \int_X h_k(x) d\mu(x).[/math]

It will usually be enough for the series to converge in some average sense to our original function.

Because polynomials and trigonometric functions tend to be particularly easy to integrate, this technique is often effective if one uses power series expansions or Fourier series expansions.
 
[PREREQUISITES]
undergraduate calculus, undergraduate real analysis

[note contributions wanted] [/note]

[EXAMPLE logint]
Suppose we want to calculate the following integral on the unit disc $\mathbb D$ of the complex plane

[math]\int_{\mathbb D} \bigg|\frac{z}{1-\bar w z}\bigg|^2 dm(w),[/math]

where $z\in\mathbb D$ and $dm$ is the Lebesgue measure in the plane. Using the geometric series expansion

[math] f(w)=\frac{z}{1-\bar w z}=z\sum_{n=0} ^\infty {\bar w}^n z^n ,[/math]

and polar coordinates we can rewrite the integral in the form

[math] \begin{align}\int_{\mathbb D} f(w) d m(w) &= \int_0 ^{2\pi} \int_0 ^1 \frac{|z|^2}{|1- z re^{-i\theta}|^2} rdr \ d\theta =|z|^2\int_0 ^{2\pi} \int_0 ^1\bigg(\sum_{n=0} ^\infty z^n r^n e^{-in\theta}\bigg)^2 rdr\ d\theta =\\&=|z|^2\int_0 ^1 \int_0^{2\pi} \sum_{n=0} ^\infty \sum_{m=0} ^\infty  z ^n \bar z^m r^{n+m} e^{i(m-n)\theta} d\theta \ rdr=\\&=|z|^2\sum_{n=0} ^\infty \sum_{m=0} ^\infty  z ^n \bar z^m \int_0 ^1 r^{n+m+1}dr  \int_0^{2\pi}e^{i(m-n)\theta} d\theta.\end{align}[/math]
There are two or three tricks in the above lines worth mentioning. First we used polar coordinates as a means of ''decoupling'' the radial and angular variable. This becomes totally apparent in the last line where we end up with a product of integrals, one involving only the radial and one involving only the angular variable. This was combined with the [[interchange integrals or sums]] trick to bring the summation operators outside the integrals. Finally observe that we used the [[square and rearrange]] trick by writing

[math]\bigg(\sum_{n=0} ^\infty z^n r^n e^{-in\theta}\bigg)^2=\sum_{n=0} ^\infty \sum_{m=0} ^\infty z^n\bar z^m r^{n+m} e^{i(m-n)\theta}.[/math]

In this case we didn't have to square first since our expression was already squared.

To finish the calculation observe that $\int_0 ^{2\pi} e^{i(m-n)\theta}d\theta$ is equal to $2\pi$ whenever $n=m$ and $0$ otherwise because of the orthogonality of the exponentials $e^{in\theta}$. Thus we have

[math]\int_{\mathbb D} \bigg|\frac{z}{1-\bar z w}\bigg|^2 dm(w)=2\pi|z|^2\sum_{n=0} ^\infty |z|^{2n} \frac{1}{2(n+1)}=\pi \log \frac{1}{1-|z|^2},[/math]

where we have used the Taylor series expansion $\log\frac{1}{1-r}=\sum_{n=1} ^\infty \frac{r^n}{n} $ for $0<r<1$. Observe that the calculation of the integral reduced to calculating integrals of power functions since we expanded our function in a power series.

[EXAMPLE]

Let us now look at a slightly more complicated variant of the integral in [ref Example #logint]. Here we have an extra logarithmic term under the integral sign

[math]\int_{\mathbb D} \bigg|\frac{z}{1-\bar w z}\bigg|^2 \log \frac{1}{1-|w|^2}\ dm(w).[/math]

Following exactly the same steps as in [ref Example #logint] we end up with the expression

[maths] \begin{align}\int_{\mathbb D} \bigg|\frac{z}{1-\bar w z}\bigg|^2 \log \frac{1}{1-|w|^2}\ dm(w)&=
2\pi|z|^2\sum_{n=0} ^\infty |z|^{2n} \int_0^1 r^{2n+1}\log\frac{1}{1-r^2} dr =\\&=\pi|z|^2\sum_{n=0} ^\infty |z|^{2n} \int_0^1 t^n\log\frac{1}{1-t} dt.\end{align}[/maths]

Now in [ref Example #logint] we ended up with the integral $\int_0^1t^n dt$ which is trivial to calculate exactly. Here we have to deal with the integral $\int_0 ^1 t^n \log \frac{1}{1-t} dt$ which does not look so trivial. However, we can once again expand the function $\log\frac{1}{1-t}$ in the power series $\sum_{k=1} ^\infty \frac{t^k}{k}$ and calculate

[math] \int_0 ^1 t^n \log \frac{1}{1-t} dt = \sum_{k=1}^\infty \frac{1}{k}\int_0^1t^{n+k}dt= \sum_{k=1}^\infty \frac{1}{k(n+k+1)}.[/math]

One can now simplify this last sum by observing that it is in fact a telescoping sum:

[math]\sum_{k=1} ^\infty\frac{1}{k(n+k+1)}=\sum_{k=0} ^\infty \frac{1}{n+1}(\frac{1}{k}-\frac{1}{n+k+1})=\frac{1}{n+1} \sum_{k=1} ^{n+1} \frac{1}{k} \simeq \frac{\log(n+1)}{n+1}.[/math]

Thus, ignoring numerical constants, our original integral can be written in the form

[math]\int_{\mathbb D} \bigg|\frac{z}{1-\bar w z}\bigg|^2 \log \frac{1}{1-|w|^2}\ dm(w)\simeq \sum_{n=1} ^\infty \frac{\log n}{n} |z|^{2n}.[/math]

One can probably look up the latter series in a table and discover that in fact

[math] \sum_{n=1} ^\infty \frac{\log n}{n} |z|^{2n} \simeq \bigg(\log\frac{1}{1-|z|^2}\bigg)^2.[/math]

However, there is a Tricki way to see that quite fast. This uses essentially the [[Divide and Conquer]] trick so we will describe this calculation in an Example therein.

[EXAMPLE] 
(prove exponential integrability of function based on the $L^p$ norms of the function.)

[GENERAL DISCUSSION]

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