How Do I Solve This Complex Contour Integral Using the Residue Theorem?

[quasar_4]

Homework Statement

Show that $$\int$$ $$\frac{ln(x) dx}{x^{3/4} (1+x)}$$ = - $$\sqrt{2}$$ Pi^2

Homework Equations

Residue theorem - integral = 2*Pi*i * sum of residues

The Attempt at a Solution

I am so lost. I don't even know where to start. I don't understand how to construct my contour. I'm not sure what I'm supposed to avoid. I don't understand how they're finding the residue. I don't get anything about this. Except that I think (!) I can show that |z*f(z)| goes to 0 as z--> 0 and --> 0 as z --> infinity. They do this in the book, but I don't really understand why (something to do with contour radii disappearing in extreme limits). I really need a better text (we have a review book that has a PAGE on this, so it's not enough, but I don't know what books are good).

All I can think of is do some kind of substitution where z = exp(i*theta), but I don't know beyond that... the examples in my book are very confusing. I really need help. My professor is out of the country for a few weeks and gave us this to work on. I don't know what to do without being able to get help

Any help, even just pointing to a good lucid text, would be so appreciated. This is a physics class, not a math class, so it would have to be reading that was clear enough to someone without complex analysis background. Thanks!

 

[mighty2000]

Hello, don't worry, I can help you with this problem. First of all, let's start with the substitution you mentioned. Let z = exp(i*theta), then dz = i*exp(i*theta) d(theta). Now we can rewrite the integral as:

\int \frac{ln(x) dx}{x^{3/4} (1+x)} = \int \frac{ln(z) dz}{z^{3/4} (1+z)}

Now, we can use the residue theorem to solve this integral. First, we need to find the poles of the function, which are the values of z that make the denominator equal to 0. In this case, we have a pole at z = -1, since 1+z = 0 when z = -1.

Next, we need to find the residues at these poles. The residue at z = -1 is given by:

Res(-1) = \frac{ln(-1) dz}{(-1)^{3/4}(1+(-1))} = \frac{ln(-1) dz}{(-1)^{3/4}(0)} = -\frac{i*\Pi}{2}

Now, we can use the residue theorem, which states that the integral of a function around a closed contour is equal to 2*Pi*i * sum of residues inside the contour. In this case, since we only have one pole inside the contour, the integral can be written as:

\int \frac{ln(z) dz}{z^{3/4} (1+z)} = 2*Pi*i * Res(-1) = 2*Pi*i * (-\frac{i*\Pi}{2}) = -\Pi^2

However, we still need to take into account the substitution we made earlier. Remember that z = exp(i*theta), so we need to change the limits of integration accordingly. When z = 0, theta = 0, and when z = \infty, theta = \Pi. Therefore, the integral can be written as:

\int \frac{ln(x) dx}{x^{3/4} (1+x)} = \int_0^\Pi \frac{ln(exp(i*theta)) i*exp(i*theta) d(theta)}{exp(i*theta)^{3/4} (1+exp(i*theta))}