-1

Here is my function $f(z)=\frac{1}{(z+1)(z+2)(z+3)...(z+100)}$ I need to find this $\int f(z)dz$ where $C:|z|=150$ counterclockwise

I tried to use second residue theorem here. So i get $\int f(z)dz=2\pi i Res_{z=0}[1/z^2 f(1/z)$ so my function here: $f(1/z)=\frac{1}{(1/z+1)(1/z+2)(1/z+3)...(z+100)}$ $1/z^2 f(1/z)=1/z^2((1/z+1)(1/z+2)(1/z+3)...(1/z+100))$

But after that i tried to use $Res_{z=0}[1/z^2 f(1/z)=lim_{z\to 0}\frac{1}{1/z^2(z+1)(z+2)(z+3)...(z+100)}$ So i get 100 times $1/0$ and that not make sense

So any what should i do to solve this integral?

Aegean
  • 99
  • Could you give or link to a statement of the second residue theorem? I'm unfamiliar with a delineation of multiple residue theorems, but haven't found anything helpful googling it. – J.G. Jan 17 '21 at 20:58
  • @J.G. By what he wrote the OP means the residue at infinity theorem, a cool instrument to deal w2ith things like this one. – DonAntonio Jan 17 '21 at 20:59
  • @DanAntonio Thanks,that seems a plausible reading. – J.G. Jan 17 '21 at 21:03

1 Answers1

0

Since $f(1/z)=\prod_{j=1}^{100}\tfrac{1}{1/z+j}=\prod_j\frac{z}{z+j}$, the correct calculation is$$z^{-2}f(1/z)=\tfrac{z^{98}}{\prod_j(z+j)}\stackrel{z\to0}{\to}0,$$making the integral $0$. As a sanity check, compare with $R>2$ to$$\oint_{|z|=R}\tfrac{dz}{(z+1)(z+2)}=\oint(\tfrac{1}{z+1}-\tfrac{1}{z+2})dz=2\pi i-2\pi i=0.$$

J.G.
  • 115,835
  • I also find 0(zero) with different way. But i suppose it has to be non-zero. So I m so confused – Aegean Jan 17 '21 at 21:22
  • @EgeTunç Could you show us the other way that gets $0$? Your question only discusses an errant approach to finding it diverges. Also, why ought it to be non-zero? – J.G. Jan 17 '21 at 21:30
  • A friend said that its answer is non-zero. Also, i supposed that the instructor of the course wouldn't be asking a zero valued question. There is not any mathematical reasoning. However, it seems $0$ – Aegean Jan 17 '21 at 21:33
  • @EgeTunç An instructor may have included it to test your understanding rather than your arithmetic. I'll give another reason it's $0$.Since $\lim_{z\to0}zf(z)=0$, $f$'s Laurent series around $0$ contains no $1/z,,1/z^2$ etc. terms.In particular, the absence of a $1/z$ term implies the result of the second residue theorem is $0$. – J.G. Jan 17 '21 at 21:35
  • okay so it has to be $0$. Thank you for detailed explanation – Aegean Jan 17 '21 at 21:40