1
$\begingroup$

I tried couple of times to compute $\int_{-\pi}^{\pi}\sin(nx)e^{inx}$. According to W.A it should be $-\pi i$, I'm losing my mind trying to understand where I got wrong.

Assuming the whole process that $e^{-in \pi}=\cos(n \pi)-i \sin(n\pi)=(-1)^{n}$, and $[\cos(nx)e^{-inx}]_{-\pi}^{\pi}$=0.

Here is what I did:

$\int_{-\pi}^{\pi}\sin(nx)e^{inx}=[\frac{(-\cos(nx))}{n}e^{-inx}]_{-\pi}^{\pi}-\int_{-\pi}^{\pi}\frac{ine^{-inx}\cos(nx)}{n}=-\frac{1}{n}[\cos(nx)e^{-inx}]_{-\pi}^{\pi}-\int_{-\pi}^{\pi}ie^{-inx}\cos(nx)e^{inx}=0-i[\frac{e^{-inx}}{-in}]_{-\pi}^{\pi}\cos(nx)=0.$

I'd really love to understand what is wrong with that.

Thank you very much.

$\endgroup$
2
  • $\begingroup$ Yeah, I used it as you can see, twice. $\endgroup$
    – Jozef
    Commented Feb 5, 2012 at 20:44
  • $\begingroup$ It should be noted that the above equality (and subsequently the answers) hold so long as $n \neq 0$. $\endgroup$
    – JavaMan
    Commented Feb 6, 2012 at 1:21

2 Answers 2

6
$\begingroup$

You should use

$$\sin(n x)=\frac{e^{inx}-e^{-inx}}{2i}.$$

Then

$$\int_{-\pi}^\pi \frac{e^{inx}-e^{-inx}}{2i} e^{inx}dx=\frac{1}{2i}\int_{-\pi}^\pi(e^{2inx}-1)dx$$

but

$$\frac{1}{2i}\int_{-\pi}^\pi e^{2inx}dx=0$$

and you are left with $i\pi$.

$\endgroup$
0
2
$\begingroup$

For what it's worth, and just for kicks, let's try to use the "integrate by parts twice and solve for the integral trick":

$$\eqalign{ \int \underbrace{\vphantom{(}e^{Bx}}_u \underbrace{\sin (Ax)\, dx}_{dv} &= \underbrace{-A^{-1}\cos(Ax)}_v \underbrace{e^{Bx}\vphantom{(}}_u -\int \underbrace{-A^{-1}\vphantom{(}}_v \underbrace{B\cos(Ax)e^{Bx}\,dx}_{du}\cr &= -A^{-1}\cos(Ax)e^{Bx}+A^{-1}B\Bigl[ \int \underbrace{e^{Bx}\vphantom{(}}_s \underbrace{ \cos(Ax)\,dx}_{dw}\Bigr] \cr &= -A^{-1}\cos(Ax) e^{Bx} +A^{-1}B\Bigl[ \underbrace{ A^{-1}\sin(Ax)}_w \underbrace{e^{Bx}\vphantom{(}}_s - \int \underbrace{A^{-1} \sin(Ax)}_w\underbrace{Be^{Bx}\vphantom{(}}_{ds} \Bigr] \cr &= -A^{-1}\cos(Ax)e^{Bx}+A^{-2}B \sin(Ax)e^{Bx} - A^{-2}B^2\int \sin(Ax)e^{Bx} . \cr } $$ Whence $$\eqalign{\int \sin (nx)e^{inx}\, dx &= -n^{-1}\cos(nx)e^{inx}+n^{-2}(in) \sin(nx)e^{inx} - n^{-2}(in)^2\int \sin(nx)e^{inx} \cr &= {-e^{inx} ( \cos(nx)- i \sin(nx)) \over n } -i^2 \int \sin(nx)e^{inx} \cr &= {-e^{inx} e^{-inx} \over n } -i^2 \int \sin(nx)e^{inx} \cr &= {-1 \over n } + \int \sin(nx)e^{inx}. } $$ So, the trick used in the real case (solving for the original integral) breaks down here. Note the above is true because there are constants attached to to general antiderivatives...

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .