i am looking for the reson why we must calculate the e in the very first step,for example $\int^{2\pi}_{0}e^{i(1-n)\theta}d\theta$=$2\pi i$ if n=1 and the equation =0 if n not equal to 1
A simple integrate question of which should be calculate first
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1What do you mean by "why we must calculate the e in the very first step"? Are you asking why $$\int_{|z|=1} z^{1-n} dz = \begin{cases} 2\pi i & \text{if } n = 2,\\0&\text{otherwise?}\end{cases}$$ – 2011-07-19
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0i mean the e term inside the integration – 2011-07-19
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2I don't understand what you're saying. – 2011-07-19
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1i think your comment is right except the book is saying that for n not equal to 1 then the equation =0 ,if n =1 then it is equal to 2 pi i – 2011-07-19
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0Are you sure the book has $z^{1-n}$ and not $z^{-n}$? – 2011-07-19
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0yes,that what the book says – 2011-07-19
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0Then I'm afraid but the book is wrong. – 2011-07-19
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0why,what's wrong?it is from elementary real and complex analysis by shilov – 2011-07-19
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1Because $$\int_{|z| = 1} \frac{1}{z}dz = 2\pi i$$ which is the case $n = 2$ if you insist having $z^{1-n}$ in the integrand. – 2011-07-19
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1I think there is some confusion here : $\int_{|z|=1} z^n dz=\int_0^{2\pi}e^{in\theta}\times ie^{i\theta}d\theta=i\int_0^{2\pi}e^{i(n+1)\theta}d\theta$, I think both of you have correct formulas... – 2011-07-19
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1@Theo The integral is $\mathrm d\vartheta$, not $\mathrm dz$ -- it's correct that it's $0$ if $n$ is an integer other than $1$, but it's $2\pi$, not $2\pi\mathrm i$ for $n=1$. – 2011-07-19
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0@joriki you must have forgotten a dollar sign somewhere! – 2011-07-19
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0@Olivier: Thanks, I was trying to correct it but my browser froze :-) Anyway, you'd already said basically what I wanted to say. – 2011-07-19
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0please see the bottom of the page of 392:http://books.google.com/books?id=UkQ8ANdjJUcC&printsec=frontcover&dq=elementary+real+complex&hl=en&ei=6q0lTrbdMMPKgQfpmd2CCA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCkQ6AEwAA#v=onepage&q&f=false – 2011-07-19
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0@Victor: I can't get that page in the Google Books preview. It might work if you link directly to the page? – 2011-07-19
1 Answers
Allow me to summarize the comments above, the answer, and the answer to the question that you actually asked (as I read it).
Firstly, we should recognize Olivier's comment that $\displaystyle\int_{|z|=1} z^n dz=\int_0^{2\pi}e^{in\theta}\times ie^{i\theta}d\theta=i\int_0^{2\pi}e^{i(n+1)\theta}d\theta$. This is a simple line integral. Most importantly, this is not the same as $\displaystyle \int_{|z| = 1} z^{1-n} dz$. So Theo and Victor are talking about different integrals. And both are correct (except for a factor of i in Victor's).
Now, as to evaluating $\displaystyle\int^{2\pi}_{0}e^{i(1-n)\theta}d\theta$. Firstly, if $n = 1$ then we are evaluating $\displaystyle \int_0 ^{2 \pi} 1$ and that is very clearly $2\pi$. But for all other values, we recall the following three things:
$$e^{it\pi} = \cos t + i \sin {t} \qquad \qquad \qquad \qquad \int_0 ^{2 \pi} \sin(t) dt = \int_0 ^{2 \pi} \cos(t) dt = 0$$
Then we quickly see that for $n \not = 1$ the value of the integral in question is zero.
I think that the original question was entirely based around why we care to break this up into cases, when n is or is not 1. Now that we know the answer, this might seem very apparent. Or it might not. The big key is that in evaluating this integral, I used that $e^{i t \pi} = \cos t + i \sin t$, but if $t = 0$ then this rings hollow. It says that $e^0 = \cos(0) + i \sin(0)$, which is still true ($1 = 1$).
So in that sense, there is no need to split it up into cases, or to look at e prior to the evaluation of the integral. But we do need to realize that when $n = 1$ we end up evaluating $\int_0 ^{2 \pi} \cos(0) dt = 2 \pi$ instead of something that goes to zero.
But on a more qualitative note, there is no reason not to evaluate it before-hand. As then we can do both cases in our heads immediately.
I hope this clears the air.