Finding the Residue of this function

[d2j2003]

Homework Statement

The original problem is to find the integral from 0 to 2\pi of \frac{1}{2\pi}e^{cosθ}cos(2θ)

Homework Equations

I have it down to two equations which I believe I have to find the residues of and add the results, multiply by 2\pii and I will have my answer.

The Attempt at a Solution

Have 1/2(∫e^{1/2(z+1/z)}*z + ∫e^{1/2(z+1/z)}*(1/z^{3})

so I believe I have to find Res(e^{1/2(z+1/z)}*z) and Res(e^{1/2(z+1/z)}*(1/z^{3})

If this is correct I really am unsure of how to find these residues..

 

[jackmell]

You can find the residue of z e^{\frac{z^2+1}{2z}} by using the Cauchy product:

ze^{\frac{z^2+1}{2z}}=z\sum_{n=0}^{\infty} \frac{z^n}{2^n n!} \sum_{n=0}^{\infty} \frac{1}{2^n n! z^n}=z\sum_{n=0}^{\infty}\sum_{k=0}^n \frac{z^k}{2^k k!} \frac{1}{2^{n-k} (n-k)! z^{n-k}}

Now that double sum, get all the z's together on one side. Then that's z^{1+k-(n-k)} right? When does that exponent equal -1? When ever 2k+2=n right? Now we only want the coefficients of 1/z for that to sum them up (the resisue is an infinite sum). So we need to sum all the k-values when n=2k+2. So form the inner sum using only the coefficients when n=2k+2 and compute:

\sum_{k=0}^{\infty} ?

 

[d2j2003]

If I do all of that I get \sum^{∞}_{k=0}\frac{1}{(k+2)!(2k+2)2^{3k+4}} as the coefficient of \frac{1}{z}

so this would be the residue right? Then I just have to do the same thing for the other residue and add them and multiply by 2\pii according to the residue theorem...

 

[jackmell]

Try to check them in Mathematica. It's not hard:

In[2]:=
NIntegrate[z*Exp[(z^2 + 1)/(2*z)]*I*Exp[I*t] /. 
   z -> Exp[I*t], {t, 0, 2*Pi}]

Out[2]=
5.551115123125783*^-16 + 0.8529277641641219*I

In[3]:=
Sum[1/(2^k*k!*(k + 2)!*2^(k + 2)), {k, 0, Infinity}]

Out[3]=
BesselI[2, 1]

In[4]:=
N[2*Pi*I*BesselI[2, 1]]

Out[4]=
0. + 0.8529277641641214*I

 

[d2j2003]

i'm sorry I'm kind of lost with the mathematica stuff (don't have the program).. but why did you integrate z*e^((z^2+1)/2z)*i*e^(it) ? I don't understand where the i*e^(it) came from at the end..

 

[d2j2003]

I understand everything you said about using the Cauchy product.. and went through all of that and then created a single sum from 0 to infinity substituting n=2k+2 to create the coefficient of the 1/z term in the sum, which I thought was the residue... where am I going wrong here?

Thanks for the help

 

[jackmell]

The residue of \displaystyle z e^{\frac{z^2+1}{2z}} is \displaystyle\sum_{k=0}^{\infty} \frac{1}{2^k k! (k+2)! 2^{k+2}}

That's what you should get when you make the substitution n=2k+2. So we can check that by numerically integrating:

\mathop\oint\limits_{|z|=1} z e^{\frac{z^2+1}{2z}}dz

Now let z=e^{it} so that dz=ie^{it}

and substitute that into the integral above to get:

\int_0^{2\pi} z e^{\frac{z^2+1}{2z}}ie^{it}dt,\quad z=e^{it}

That's what I coded in Mathematica.

 

[d2j2003]

ahh ok I see now.. so I did the same for the other residue (with 1/z^3 instead of z) and found that the sum is 1/z when n=2k-2 so substituted and got \sum^{∞}_{k=0}\frac{1}{k!(k-2)!2^{k-1}}

then added the two residues and multiplied by 1/2 to get \sum^{∞}_{k=0}\frac{1}{k!(k-2)!2^{k}}

(multiplied by \frac{1}{2} because the original integral had \frac{1}{4*\pi*i} in front before finding residues so \frac{1}{4*\pi*i} * 2\pi*i = \frac{1}{2}

I think this is correct.. Thanks again for your help