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[SOLVED] primitive e^{z}/z

dwsmith

Well-known member
Feb 1, 2012
1,673
Gamma is the unit circle oriented counterclockwise.

$\int_{\gamma}\frac{e^z}{z}dz$

Instead of finding the power series to solve the integral, I am trying to do it b use of the primitive.
However, I seem to not be good at finding the primitives. How can I find the primitive for this?
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
The primitive/antiderivative is not elementary. But regardless, $\frac{e^{z}}{z}$ is not analytic at $z=0$, so the fundamental theorem of calculus for analytic functions does not apply.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
The primitive/antiderivative is not elementary. But regardless, $\frac{e^{z}}{z}$ is not analytic at $z=0$, so the fundamental theorem of calculus for analytic functions does not apply.
How can I solve this integral without the use of Cauchy Integral Formula or by a power series?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,189
How can I solve this integral without the use of Cauchy Integral Formula or by a power series?
If you know residue theory, you could apply that.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
If you know residue theory, you could apply that.
If we re-write the integral as,

$$
i\int_0^{2\pi}e^{e^{it}}dt
$$

we would have the above integral. I am not sure if that makes it any easier.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,189
If we re-write the integral as,

$$
i\int_0^{2\pi}e^{e^{it}}dt
$$

we would have the above integral. I am not sure if that makes it any easier.
Hmm. Just trying to reproduce that expression. If you let $z=e^{it}$, then $dz=ie^{it}\,dt$, and the integral becomes
$$\int_{0}^{2\pi}\frac{e^{e^{it}}}{e^{it}}\,ie^{it}\,dt=i\int_{0}^{2\pi}e^{e^{it}}\,dt,$$
as you say. However, that integral has no elementary antiderivative.

I would probably use residues to compute this integral, but, if you aren't to that point in your course yet, you're back to something like the Cauchy Integral Formula, which I know you said you didn't want to use. Is that because you're not allowed to use it, or because you don't want to use it?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Hmm. Just trying to reproduce that expression. If you let $z=e^{it}$, then $dz=ie^{it}\,dt$, and the integral becomes
$$\int_{0}^{2\pi}\frac{e^{e^{it}}}{e^{it}}\,ie^{it}\,dt=i\int_{0}^{2\pi}e^{e^{it}}\,dt,$$
as you say. However, that integral has no elementary antiderivative.

I would probably use residues to compute this integral, but, if you aren't to that point in your course yet, you're back to something like the Cauchy Integral Formula, which I know you said you didn't want to use. Is that because you're not allowed to use it, or because you don't want to use it?
We have learned it but we are supposed to try this without those methods so we will have more appreciation for them.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,189
We have learned it but we are supposed to try this without those methods so we will have more appreciation for them.
Well, what methods are you allowed to use? As RV said above, the Fundamental Theorem is inapplicable here. I suppose you could try to fiddle around with other parametrizations, and see if you can dream up one that has an antiderivative. What if you tried something like

$$z=e^{if(t)}\implies dz=i f'(t) e^{if(t)}\,dt.$$
Then you'd get
$$\int_{\gamma}\frac{e^{z}}{z}\,dz=\int_{a}^{b} \frac{ e^{e^{if(t)}}}{e^{if(t)}}\,i f'(t) e^{if(t)}\,dt=\int_{a}^{b} e^{e^{if(t)}}\,i f'(t) \,dt=i\int_{a}^{b} e^{e^{if(t)}}\,f'(t) \,dt.$$
Now, what we would like is for
$$\frac{d}{dt}e^{if(t)}=if'(t),$$
because then we could use the chain rule. Can we solve this DE? Well, we compute the LHS thus:
$$i f'(t)e^{if(t)}=if'(t).$$
If $f'(t)=0$, then we get $f(t)=\text{const}$, which will not produce a valid parametrization. So, if we cancel we get $e^{if(t)}=0$. This equation has no solutions.

Therefore, this kind of parametrization will not yield an integral that will succumb to a substitution.

What about by-parts? With the previous substitution, we arrived at
$$i\int_{a}^{b} e^{e^{if(t)}}\,f'(t) \,dt.$$
By-parts yields
$$i\left[\int_{a}^{b} \underbrace{e^{e^{if(t)}}}_{u}\,\underbrace{f'(t) \,dt}_{dv}\right]=i\left[e^{e^{if(t)}}f(t)\Big|_{a}^{b}-\int_{a}^{b} e^{e^{if(t)}}e^{if(t)}if'(t)\,f(t) \,dt\right].$$
Now let's try again for a substitution. We would like
$$\frac{d}{dt}(e^{if(t)}+if(t))=if'(t)f(t).$$
Performing the indicated differentiation yields
$$e^{if(t)}if'(t)+if'(t)=if'(t)f(t).$$
The option $f'(t)=0$ isn't a good option, again, so we cancel:
$$e^{if(t)}+1=f(t).$$
You can solve for $f(t)$ in terms of the product log function, but the result is a constant again.

Gotta go. Maybe these failed attempts will suggest something to you.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
I think I might have done this in a previous post, but I'll do it again.

$\displaystyle \int_{\gamma} \frac{e^{z}}{z} \ dz = i \int_{0}^{2 \pi} e^{e^{it}} \ dt $

$ = \displaystyle i \int_{0}^{2 \pi} \left(1 + e^{it} + \frac{e^{2it}}{2!} + \ldots \right) $

$= \displaystyle i \int_{0}^{2 \pi} \left(1+ \cos(t) + i \sin(t) + \frac{\cos(2t)}{2!} + i \frac{\sin(2t)}{2!} + \ldots \right)$

Now integrate termwise.

The first integral integral evaluates to $ 2 \pi i$. All other integrals evaluate to zero.