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

2.6 Q22

(1/3) > >>

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.

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.


[0] Message Index

[#] Next page

Go to full version