1. ## cauchy integral formula

Eval

$\displaystyle \frac{1}{2\pi i}\oint\frac{e^{tz}}{(z^2+1)^2}dz$ if t>0 and C is the circle {|z|=3}

Hmm...

Am I suppose to parameterise something?

I can write for example z(t)=3e^{it} then dz = 3i e^{it}dt and sub it back in, but that just makes it messy and complicated.

Am I missing an easier way?

2. Originally Posted by scorpion007
Eval

$\displaystyle \frac{1}{2\pi i}\oint\frac{e^{tz}}{(z^2+1)^2}dz$ if t>0 and C is the circle {|z|=3}

Hmm...

Am I suppose to parameterise something?

I can write for example z(t)=3e^{it} then dz = 3i e^{it}dt and sub it back in, but that just makes it messy and complicated.

Am I missing an easier way?
Do you know the Residue theorem? You're function has singularities at $\displaystyle z=\pm i$ both of which lie in $\displaystyle C$. So, $\displaystyle \oint_{C}f(z)dz=\underset{z= i}{\text{Res}}f(z)+\underset{z=- i}{\text{Res}}f(z)$

3. Thanks, I considered that just now, but this question belongs to a section well-prior to covering the residue thm, in the CIF section. So I suppose it should be solvable using that...

4. Originally Posted by scorpion007
Thanks, I considered that just now, but this question belongs to a section well-prior to covering the residue thm, in the CIF section. So I suppose it should be solvable using that...
Ok, decompose your the denominator into $\displaystyle (z-i)^2(z+i)^2$ and decompose this using partial fractions and then from there apply Cauchy's integral theorem.

5. I get

$\displaystyle -\frac{t}{2} \cos t$

but the answer says: $\displaystyle 1/2 (\sin t -t \cos t)$

for the partial fractions I got A = B = -1/4 e^{tz}

so we have

$\displaystyle \frac{1}{2\pi i} \oint_C\frac{e^{tz}}{(z^2+1)^2}dz=$

$\displaystyle \frac{1}{2\pi i} \left\{-\left(\oint_C\frac{e^{tz}/4}{(z+i)^2}\; dz + \oint_C\frac{e^{tz}/4}{(z-i)^2}\;dz\right)\right\}=$

$\displaystyle -\frac{1}{2\pi i}2\pi i \frac{1}{4}(f'(-i)+f'(i))$ where

$\displaystyle f(z)=e^{tz}$ and $\displaystyle f'(z)=te^{tz}$

$\displaystyle =-\frac{t}{4}(e^{-it}+e^{it})$

$\displaystyle =-\frac{t}{4}(2\cos t)$

$\displaystyle =-\frac{t}{2}\cos t$

Did I make a mistake?

6. Originally Posted by scorpion007
I get

$\displaystyle -\frac{t}{2} \cos t$

but the answer says: $\displaystyle 1/2 (\sin t -t \cos t)$

for the partial fractions I got A = B = -1/4 e^{tz}

so we have

$\displaystyle \frac{1}{2\pi i} \oint_C\frac{e^{tz}}{(z^2+1)^2}dz=$

$\displaystyle \frac{1}{2\pi i} \left\{-\left(\oint_C\frac{e^{tz}/4}{(z+i)^2}\; dz + \oint_C\frac{e^{tz}/4}{(z-i)^2}\;dz\right)\right\}=$

$\displaystyle -\frac{1}{2\pi i}2\pi i \frac{1}{4}(f'(-i)+f'(i))$ where

$\displaystyle f(z)=e^{tz}$ and $\displaystyle f'(z)=te^{tz}$

$\displaystyle =-\frac{t}{4}(e^{-it}+e^{it})$

$\displaystyle =-\frac{t}{4}(2\cos t)$

$\displaystyle =-\frac{t}{2}\cos t$

Did I make a mistake?
I don't think you decomposed it correct.

7. ah yes! thanks!