# How to justify interchange of sum and integral in fourier series?

$f$ is the $4$-periodic function $f(x) = 1$ if $x \in [0,2)$ and $f(x) = - 1$ if $x \in [2,4)$

The Fourier series of $f$ is

$$F(t) = {4 \over \pi} \sum_{n=1}^{\infty} {\sin({\pi \over 2}(2n + 1)t) \over 2n + 1}$$

for sure because I calculated it like three times

I want to find the sum $\sum_{n = 1}^{\infty} 1/n^2$ using this. I know I should integrate from $0$ to $1$:

$$1 = \int_0^1 f(t) dt = \int_0^1 F(t) dt = {4 \over \pi} \sum_{n=1}^{\infty} {1 \over 2n+1}\int_0^1 \sin({\pi \over 2}(2n + 1)t)dt = {8 \over \pi^2} \sum_{n=1}^{\infty} {1 \over (2n + 1)^2}$$

now I just break the sum into even and odd terms to find its value

question: how to give complete justification for this interchange of sum and integral?

Help is much appreciated

we can't interchange integral and sum unless the series is uniformly convergent or the function is positive.. maybe there are weaker conditions but this does not always work!

• Find in a textbook on Fourier series: the Fourier series of a piecewise $C^1$ function converges to the function wherever it is continuous. – GEdgar Jun 13 '16 at 20:42
• And at jumps the Fourier series converges to the average of the two one-sided limits. – GEdgar Jun 13 '16 at 20:48

The series is not uniformly convergent on an interval that includes $0$, typical of Fourier series converging to piecewise continuous functions with jump discontinuities. Furthermore, it is not always possible to apply dominated convergence to justify interchanging the sum and integral.

Nevertheless, you can apply a general result that the Fourier series of a function in $L^1$ can be integrated term-by-term.

Suppose, in general, that $f$ is $2T-$periodic with Fourier sine series

$$f(t) = \sum_{n=1}^\infty b_n \sin\frac{n \pi t}{T}.$$

The series obtained formally by integrating term-by term 0ver $[0,t]$ is

$$\hat{F}(t) = -\sum_{n=1}^\infty \frac{Tb_n}{\pi n} \cos\frac{n \pi t}{T} + \sum_{n=1}^\infty\frac{Tb_n}{\pi n},$$

and the series converges by the comparison test.

Now consider

$$F(t) = \int_0^t f(s) \, ds.$$

Since $f$ is integrable and periodic, it follows that $F$ is absolutely continuous with the same periodicity. Hence, $F$ has a convergent Fourier series

$$F(t) = \frac{A_0}{2} + \sum_{n=1}^\infty \left( A_n\cos\frac{n \pi t}{T} + B_n \sin\frac{n \pi t}{T} \right),$$

where

$$A_n = \frac{1}{T}\int_0^{2T}F(t) \cos \frac{n \pi t}{T} \, dt, \\ B_n = \frac{1}{T}\int_0^{2T}F(t) \sin \frac{n \pi t}{T} \, dt.$$

Now use integration by parts, with boundary terms vanishing due to periodicity , to obtain

$$A_n = \left.f(t) \frac{1}{\pi}\sin \frac{n \pi t}{T}\right|_0^{2T} - \frac{1}{T} \int_0^{2T} f(t) \frac{T}{\pi n} \sin \frac{n \pi t}{T} \, dt = - \frac{T b_n}{\pi n}, \\ B_n = \left.-f(t) \frac{1}{\pi}\cos \frac{n \pi t}{T}\right|_0^{2T} + \frac{1}{T} \int_0^{2T} f(t) \frac{T}{\pi n} \cos \frac{n \pi t}{T} \, dt = 0.$$

Hence,

$$F(t) = \frac{A_0}{2} - \sum_{n=1}^\infty \frac{Tb_n}{\pi n}\cos\frac{n \pi t}{T}.$$

Since, $F(0) = 0$ we have

$$\frac{A_0}{2} = \sum_{n=1}^\infty \frac{Tb_n}{\pi n},$$

and

$$F(t) = \hat{F}(t).$$

• I thought that only infinite sum and infinite integral are not always inter-changbale. Finite integral and infinite integral are always interchangeable. So are finite sun and infinite interchangeable. – mike Jun 19 '16 at 9:24