Prove that $ I = \int_{0}^1 \sin(2 \pi x) e^{-2 \pi ikx} dx = \begin{cases} \frac{-ki}{2}, & \text{if $k=\pm1$ } \\ 0, & \text{if $k=0$} \end{cases}$

Question

Prove that $$ I = \int_{0}^1 \sin(2 \pi x) e^{-2 \pi ikx} dx = \begin{cases} \frac{-ki}{2}, & \text{if $k=\pm1$ } \\ 0, & \text{if $k=0$} \end{cases}$$

According to our definition of Fourier Series $\hat{f}(k) = \int_{0}^1 f(x) e^{-2 \pi ikx} dx $ is the k-th Fourier coefficient of $f$.

For $k=0$ I get $I=0$, since the $$\int_{0}^1 \sin(2 \pi x)e^{0}= 0$$

For $k=\pm1$ I am struggling.

first $k=1$, $$\int_{0}^1 \sin(2 \pi x) e^{-2 \pi ikx} dx$$ I use integration by parts and $f=\sin(2 \pi x)$, $f^\prime(x)= \cos(2 \pi x)2 \pi x$, $g=\frac{e^{-2 \pi ikx}}{-2 \pi ikx}$ and $g^\prime(x) = e^{-2 \pi ikx}$

$$ \left[ \sin(2 \pi x) \frac{e^{-2 \pi ikx}}{-2 \pi ik} \right]_0^{1} - \int_{0}^1 \cos(2 \pi x)2\pi \frac{e^{-2 \pi ikx}}{-2 \pi ik} dx $$ First term is zero when evaluating at $x=0$ and $x=1$ and integrating again by parts

$$ = \frac{-2\pi}{-2 \pi ik} \int_{0}^1 \cos(2 \pi x) e^{-2 \pi ikx} dx$$ $$\frac{1}{ik} \left[ \left[ \cos(2 \pi x)2\pi \frac{e^{-2 \pi ikx}}{-2 \pi ik} \right]_0^{1} -\int_{0}^1 -\sin(2 \pi x)2 \pi \frac{e^{-2 \pi ikx}}{-2 \pi ikx} dx\right] $$

where i had $f=\cos(2 \pi x)$, $f^\prime(x)= -\sin(2 \pi x)2 \pi x$, $g=\frac{e^{-2 \pi ikx}}{-2 \pi ikx}$ and $g^\prime(x) = e^{-2 \pi ikx}$ and used that $\frac{1}{i}=-i$

I get the $I$ again

$$ I = \frac{1}{ik} \frac{1}{-ik} I$$ For $k=-1$ It is similar with one sign change. I don't have a clue how to proceed. Any help appreciated! Thanks!

Answer 1

$$ I = \int_{0}^1 \sin(2 \pi x) e^{-2 \pi ikx} dx=\dfrac1{2 i}\int_0^1(e^{2\pi ix}-e^{-2\pi ix}) e^{-2 \pi ikx} dx\\ =\dfrac1{2 i}\int_0^1(e^{2\pi ix}-e^{-2\pi ix}) e^{-2 \pi ikx} dx =\dfrac1{2 i}\int_0^1(e^{2\pi i(1-k)x}- e^{-2 \pi i(1+k)x}) dx\\ $$ For $k=0$ it is solved by the author above. For $k=1$ we get: $$ I=\dfrac1{2 i}\int_0^1(1- e^{-4 \pi ix}) dx=\dfrac1{2i}[\int_0^1(1-\cos4\pi x+i\sin4\pi x)dx $$ The integral of the last 2 terms in an integral over periodic even functions over twice its period and therfore integrates to zero. Thus $$ I =\dfrac1{2i}[\int_0^1dx=-\dfrac i2 $$ Similarly we get for $k=-1$: $$ I=\dfrac1{2 i}\int_0^1( e^{ 4 \pi ix}-1) dx=\dfrac1{2i}\int_0^1[( \cos4\pi x +i\sin4\pi x)-1]dx=\dfrac i2 $$ We can combine both results for the case $k=\pm1$ and write in this case: $$ I=-\dfrac{ik}2 $$