0
$\begingroup$

Calculate $\int{ \frac{e^z}{z^2(z^2+3)}}dz$ over the rectangle $x=2,x=-2 ,y=2,y=-2$.

What i did is find the roots of $z^2+3$ break the $\frac{1}{z^2+3}$ into $\frac{-i/6}{z-\sqrt{3}i} +\frac{i/6}{z+\sqrt{3} i}$ and the use cauchys integral formula for the function $e^zz^{-2}$ two times one for $-3i$ and for $3i$ my final result was $πi/9$. I need a full proof please so to understand the method.Without using residues.!!

$\endgroup$
2
  • $\begingroup$ $z^2 + 3 = (z+\sqrt3i)(z-\sqrt3i)$, not $(z+3i)(z-3i)$. $\endgroup$
    – user307169
    Aug 18, 2016 at 16:31
  • 1
    $\begingroup$ @tilper ye of course typing mistake $\endgroup$
    – Jam
    Aug 18, 2016 at 16:34

1 Answer 1

5
+50
$\begingroup$

You approach was wrong because you used Cauchy formula with a function which is not holomorphic in the considered domain.

Considering the rectangle over which you integrate, you can see that all the poles of your function : $0,i\sqrt{3},-i\sqrt{3}$ are inide the considered domain, so applying Cauchy formula does not seem to be a good idea.

But you can apply Residue theorem :

  • $i\sqrt{3}$ is a simple pole : $$Res\left(\frac{e^z}{z^2(z^2+3)},i\sqrt{3}\right)=\lim_{z \to i\sqrt{3}}\frac{e^z}{z^2(z+i\sqrt{3})}=\frac{-e^{i\sqrt{3}}}{6i\sqrt{3}}$$
  • $-i\sqrt{3}$ is a simple pole : $$Res\left(\frac{e^z}{z^2(z^2+3)},-i\sqrt{3}\right)=\lim_{z \to -i\sqrt{3}}\frac{e^z}{z^2(z-i\sqrt{3})}=\frac{e^{-i\sqrt{3}}}{6i\sqrt{3}}$$
  • $0$ is a pole of order $2$ : $$Res\left(\frac{e^z}{z^2(z^2+3)},0\right)=\lim_{z \to 0}\left(\frac{e^z}{z^2+3}\right)'=\frac{1}{3}$$

So we apply Residue theorem :

$$\int{ \frac{e^z}{z^2(z^2+3)}}dz=2\pi i\left(\frac{-e^{i\sqrt{3}}}{6i\sqrt{3}} +\frac{e^{-i\sqrt{3}}}{6i\sqrt{3}}+\frac{1}{3}\right)=2\pi i\left(\frac{\sin(-\sqrt{3})}{3\sqrt{3}}+\frac{1}{3} \right)$$


Without Residues you can write : $$\frac{e^z}{z^2(z^2+3)}=\frac{ae^z}{z^2}+\frac{be^z}{z}+\frac{ce^z}{z+i\sqrt3}+\frac{de^z}{z-i\sqrt3}$$ You solve the system induced by this equality and get : $$\frac{e^z}{z^2(z^2+3)}=\frac{\frac{1}{3}e^z}{z^2}+\frac{0e^z}{z}+\frac{\frac{1}{i6\sqrt3}e^z}{z+i\sqrt3}+\frac{-\frac{1}{i6\sqrt3}e^z}{z-i\sqrt3}$$

$$\int{ \frac{e^z}{z^2(z^2+3)}}dz=\int{\frac{\frac{1}{3}e^z}{z^2}}dz+\int{\frac{\frac{1}{i6\sqrt3}e^z}{z+i\sqrt3}}dz+\int{\frac{-\frac{1}{i6\sqrt3}e^z}{z-i\sqrt3}}dz$$ Then you can calculate each of these integrals with Cauchy theorem. $$\int{\frac{\frac{1}{3}e^z}{z^2}}dz=\frac{1}{3}\int{\frac{e^z}{z^2}}dz=\frac{2\pi i}{3}e^0=\frac{2\pi i}{3}$$ Here I used the formula $f'(a)=\frac{1}{2\pi i}\int\frac{f(z)}{(z-a)^2}$ (with the sames conditions than the classic Cauchy formula). $$\int{\frac{\frac{1}{i6\sqrt3}e^z}{z+i\sqrt3}}dz=\frac{1}{i6\sqrt3}\int{\frac{e^z}{z+i\sqrt3}}dz=\frac{2\pi ie^{-i\sqrt{3}}}{6i\sqrt{3}}$$ $$\int{\frac{-\frac{1}{i6\sqrt3}e^z}{z-i\sqrt3}}dz=\frac{2\pi i(-e^{i\sqrt{3}})}{6i\sqrt{3}}$$ Finally : $$\int{ \frac{e^z}{z^2(z^2+3)}}dz=2\pi i\left(\frac{-e^{i\sqrt{3}}}{6i\sqrt{3}} +\frac{e^{-i\sqrt{3}}}{6i\sqrt{3}}+\frac{1}{3}\right)=2\pi i\left(\frac{\sin(-\sqrt{3})}{3\sqrt{3}}+\frac{1}{3} \right)$$

$\endgroup$
3
  • $\begingroup$ i need an answer without using residues! $\endgroup$
    – Jam
    Aug 19, 2016 at 10:35
  • 1
    $\begingroup$ You can replace each residue calculation with Cauchy's integral formula. For example $$\int_{|z|=r} \frac{e^z}{z^2(z^2+3)} \, \rm{d} z = 2\pi i \left( \frac{e^z}{z^2+3}\right)^\prime_{z=0} = 2 \pi i \left( \frac{e^z(3-2z+z^2)}{(z^2+3)^2} \right)_{z=0} = \frac{2\pi i}{3},$$ where $r>0$ is sufficiently small so that $|z|\le r$ contains only the singularity at $z=0$. $\endgroup$ Aug 19, 2016 at 20:12
  • $\begingroup$ @ManolisLyviakis I added an answer without residues. $\endgroup$
    – Bérénice
    Aug 19, 2016 at 23:05

You must log in to answer this question.

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