MAT334-2018F > MAT334--Lectures & Home Assignments

2.6 Q22

**Kathy Ngo**:

Can someone show me how they did (a)?

I have no idea how I'm supposed to use (8 ) or (9) to solve.

**Ende Jin**:

I don't understand your question.

Isn't it just substituting in the number?

**Min Gyu Woo**:

The steps of getting to that final equation. Not the formula.

**ruienlin**:

Hi Kathy, here is my solution to (a):

For Question(a),$$ \int_0^\infty \frac{d_x}{8+x^3}=\frac{1}{8} \int_0^\infty \frac{d_x}{1+\frac{x^3}{8}}$$

$$\int_0^\infty \frac{d_x}{1+\frac{x^3}{8}}=\int_0^\infty \frac{d_x}{1+ ( \frac{x}{2})^3} $$

Change variable $x/2$ to $t$ and apply (9), then it becomes

$$2\int_0^\infty \frac{d_t}{1+ t^3} = \frac{2}{3}\frac{\pi}{\sin(\frac{\pi}{3})}=\frac{4\sqrt{3}\pi}{9}$$

so,$$ \int_0^\infty \frac{d_x}{8+x^3}=\frac{\sqrt{3}\pi}{18}$$

**Victor Ivrii**:

consider (c), which is the most general, and consider a "pizza contour" $\gamma$ with an angle $\theta$ such that $e^{i\beta\theta}=1$ (it may be greater than $2\pi$, but it really does not matter) and also if $-1<\gamma <0$ you should cut a small piece of the radius $\varepsilon$. Consider

$$

\int_\gamma \frac{z^\gamma\,dz}{1+z^\beta}.

$$

What are singularities inside? Calculate the residue (or residues).

Prove that for $R\to \infty$ the integral over big arc tends to $0$

Prove that for $\varepsilon\to 0$ the integral over big arc tends to $0$

Express integrals over straight segments via

$$\int_\varepsilon ^R \frac{x^\gamma\,dx}{1+x^\beta}.$$

Then after taking the limits you'll be able to find integral in question.

There are Examples in the section, using exactly the same method.

Navigation

[0] Message Index

[#] Next page

Go to full version