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I'm working on tough integrals that basically contain a fraction inside them. Here's a (simplified) example:

$$\int_{-\pi}^\pi{\frac{1+e^{i t}}{e^{i t}}dt}$$

I'm interested in solving this using differentiation under the integral, and I'm hoping that someone can help me. My work so far follows...

First, I restate the integral as follows:

$$-i\int_{-\pi}^\pi{\frac{1+e^{i t}}{e^{i t}}\frac{ie^{i t}}{e^{i t}}dt}= -i\int_{|z|=1}{\frac{1+z}{z}\frac{1}{z}dz}$$

Here's where we can use differentiation under the integral, even if it seems too much for such a simple problem.

The integral is now: $$-i\int_{|z|=1}{\frac{1+z}{z^2}dz}$$

and the fractions in some of the problems that I'm working on will be extremely hard to work with. So I set up a new function:


Note that taking the derivative w.r.t. $y$ yields the result I'm looking for, if $y$ is then set equal to $1$:

$$F'(y)=-i\int_{|z|=1}{\frac{1+z}{z^2}\cos{\left(\frac{y}{z^2}-\frac{1}{z^2}\right)}dz} $$


This method then seems extremely easy to do, even with far more complicated divisors in the fractions, since I can use Cauchy to integrate, and then proceed with differentiation under the integral. I'd like to know if this seems correct so far. I'm wondering if there are additional considerations I need to be aware of before getting too complex. But it seems that I have a chance of "building" integrals in this way, so I want to be sure that I'm correct.

If this seems correct and checks out, I'd really like to know, and maybe this is too much for this question, if I can use this with multiple fractions and multiple variables. For instance, would a function like $\int{\frac{p(x)}{q(x)r(x)}}$ be easy to adapt to this method?

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I don't see what you have gained by rewriting your integral as $-i\int (1+z) \sin(\frac y{z^2} - \frac1{z^2})\; dz$? Am I misunderstanding something here? The second integral seems much more complicated than the first. –  Sam Sep 1 '11 at 19:03
@Sam: The example above is supposed to be a simple example. I'm working with extremely tough integrals, and they're tough because of the fractions. See for a more representative example. Essentially, I believe that I can eliminate the fraction on the bottom and replace the integral with something that is easy to work with, since Cauchy and complex analysis can be used to quickly find the new, easier integral. They are hard to integrate otherwise. –  Matt Groff Sep 1 '11 at 20:17

1 Answer 1

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You state that your integral is equal to $-i\int_{|z|=1}\frac{1+z}{z}\frac{1}{z}\mathrm{d}z$. This is $-i\int_{|z|=1}\left(\frac{1}{z^2}+\frac{1}{z}\right)\mathrm{d}z$. Since the residue of $z^{-2}$ is $0$, and the residue of $z^{-1}$ is $1$, we get that $$ -i\int_{|z|=1}\left(\frac{1}{z^2}+\frac{1}{z}\right)\mathrm{d}z=-i(0+2\pi i)=2\pi $$ Another way of evaluating your integral is $$ \begin{align} \int_{-\pi}^\pi\frac{1+e^{it}}{e^{it}}\mathrm{d}t &=\int_{-\pi}^\pi\left(1+e^{-it}\right)\mathrm{d}t\\ &=2\pi+0 \end{align} $$ Since the integral of $e^{-it}$ over a complete period is $0$.

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