# Evaluate real integral using residue theorem, where did I go wrong?

• Poopsilon
In summary, the integral \int_0^{\pi} \frac{dt}{(a+cost)^2} for a > 1 can be evaluated using the residue theorem. After simplification and taking the limit, the correct answer is \frac{\pi a}{(a^2 - 1)\sqrt{a^2 - 1}}. The mistake made was using -b instead of b in the calculations.
Poopsilon
Edit: Never mind I found my error, moderator can lock this.

## Homework Statement

Evaluate the integral $\int_0^{\pi} \frac{dt}{(a+cost)^2}$ for a > 1.

## Homework Equations

$$\int_0^{\pi}\frac{dt}{(a+cost)^2} = \pi i\sum_{a\epsilon \mathbb{E}}Res(f;\alpha)$$

Where $\mathbb{E}$ is the open unit disk, and $f(z) = \frac{1}{iz(a+\frac{1}{2}(z+\frac{1}{z}))^2}$.

## The Attempt at a Solution

$$f(z) = \frac{1}{iz(a+\frac{1}{2}(z+\frac{1}{z}))^2} = \frac{1}{iz\frac{(z^2 + 2az + 1)^2}{4z^2}} = \frac{-4iz}{(z^2 + 2az + 1)^2} = \frac{-4iz}{[(z + a + \sqrt{a^2 - 1})(z + a - \sqrt{a^2 - 1})]^2}$$

Thus f has two roots of multiplicity two. It's fairly easy to see that only the root $z = -a + \sqrt{a^2 - 1}$ lies inside the open unit disk for a > 1, thus we set $b = -a + \sqrt{a^2 - 1}$ and obtain:

$$\pi i\sum_{\alpha\epsilon\mathbb{E}}Res(f;\alpha) = Res(f;b) = lim_{z\rightarrow b}\frac{d}{dz}\frac{-4iz(z+a-\sqrt{a^2 - 1})^2}{(z + a + \sqrt{a^2 - 1})^2(z + a - \sqrt{a^2 - 1})^2} = lim_{z\rightarrow b}4\pi\frac{d}{dz}\frac{z}{(z+a+\sqrt{a^2 - 1})^2} = lim_{z\rightarrow b}4\pi \frac{a + \sqrt{a^2 - 1} -z}{(z + a + \sqrt{a^2 - 1})^3}$$

From here I take the limit and after simplification obtain $\frac{\pi\sqrt{a^2 - 1}}{a^3}$. Yet the book says my answer should be $\frac{\pi a}{(a^2 - 1)\sqrt{a^2 - 1}}$. Where have I gone wrong? Thanks.

Last edited:
Explain your mistake so that everyone can learn from it!

It wasn't really a mistake in my understanding of the theory it was just a sign error. I was accidentally plugging in -b instead of b.

## 1. What is the residue theorem?

The residue theorem is a mathematical tool used to evaluate complex integrals. It states that for a function with a pole of order n at a point z0, the integral around a closed contour C can be evaluated by summing the residues of the function at all the poles inside the contour.

## 2. How do I use the residue theorem to evaluate a real integral?

To use the residue theorem to evaluate a real integral, you first need to transform the integral into a complex integral. This can be done by extending the limits of integration to include the pole(s) of the integrand. Then, you can use the residue theorem to evaluate the complex integral and take the real part as the value of the original integral.

## 3. Where do I need to be careful when using the residue theorem?

When using the residue theorem, you need to be careful to correctly identify all the poles inside the contour and calculate their corresponding residues accurately. You also need to make sure that the contour used does not cross any of the poles, as this can affect the value of the integral.

## 4. What are some common mistakes when using the residue theorem to evaluate integrals?

Some common mistakes when using the residue theorem include miscalculating the residues, using the wrong contour, or forgetting to consider all the poles inside the contour. It is also important to remember to take the real part of the complex integral to get the final value of the real integral.

## 5. Where did I go wrong if my calculated value using the residue theorem does not match the actual value of the integral?

If your calculated value using the residue theorem does not match the actual value of the integral, it is likely that you made a mistake in identifying the poles or calculating the residues. It is also possible that the chosen contour was not appropriate for the integral, or the integral itself is not suitable for evaluation using the residue theorem.

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