Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Just wondering if this is right, or the right approach as I've not done it before!

Evaluate: $$\int_{|z|=1} \frac {\sin(z)}z dz$$

I expanded $\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots$

So $\frac{\sin(z)}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots$ Which left an integral of: $$\int_{|z|=1}\sum_{n=0}^{\infty} \frac{(-1)^nz^{2n}}{(n+1)!} dz$$

From the unit circle, $z=e^{i\theta} \Rightarrow dz = ie^{i\theta} d\theta$ which gives:

$$\sum_{n=0}^{\infty}\frac {(-1)^ni}{(n+1)!}\int_{0}^{2\pi} e^{3ni\theta} d\theta$$ $$ = \sum_{n=0}^{\infty} \frac {(-1)^n(e^{6\pi ni}-1)}{3n(n+1)!} $$ $$ = \sum_{n=1}^{\infty} \frac {-2(-1)^n}{3n(n+1)!} $$

Bit unsure if that's right, or the best approach if it is. Thanks!

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Most of the work looks great.

But note that $e^{3ni2\pi} = e^ {6\pi n i} = 1$, not $-1$, unless $n=0$, and that $e^{3ni(0)} = 1$ always. So the answer is much simpler than what you written in the end.

share|improve this answer
    
Oh, so it's just zero! Thank you very much. –  Mike Miller Mar 18 '13 at 10:07
    
@Brit: You should take a look at the $n=0$ term a little bit closer. –  mixedmath Mar 18 '13 at 10:09
    
I thought it would be 0/0? –  Mike Miller Mar 18 '13 at 10:10
    
@Brit Integrating $e^{0}$ from $0$ to $2\pi$ does not give zero, and $(-1)^0i/(0+1)!$ is also not zero. So you oversimplified your integral a little bit. –  mixedmath Mar 18 '13 at 10:11
1  
Oh, I see - staring me in the face! So simply $2\pi i$ –  Mike Miller Mar 18 '13 at 10:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.