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.

[yuruoyun]

For Question(b),$$ \int_0^\infty \frac{xd_x}{x^4+16} = \frac{1}{16} \int_0^\infty \frac{xd_x}{\frac{x^4}{16}+1}$$
Change variable $x/2$ to $t$ and apply 8, then it becomes
$$2\int_0^\infty \frac{2td_t}{1+ t^4} = 4\int_0^\infty \frac{td_t}{1+ t^4}$$
$\beta = 4, \alpha\beta-1 = 1$, so $\alpha = \frac{1}{2}$
$$\int_0^\infty \frac{xd_x}{\frac{x^4}{16}+1}=\frac{4}{4}\frac{\pi}{\sin(\frac{\pi}{2})}=\pi$$
so$$\int_0^\infty \frac{xd_x}{x^4+16} = \frac{\pi}{16} $$
Question(c) is the same following the same logic, with $0\leq\gamma\lt\beta-1$
$$\int_0^\infty \frac{x^\gamma d_x}{1+x^\beta}$$
$$\alpha\beta-1=\gamma$$
$$\alpha=\frac{\gamma+1}{\beta}, \frac{1}{\beta}\le\frac{\gamma+1}{\beta}\lt1$$
$$\int_0^\infty \frac{x^\gamma d_x}{1+x^\beta} = \frac{\pi}{\beta\sin(\frac{(\gamma+1)\pi}{\beta})}$$

[Victor Ivrii]

You should not use ready formulae but to derive the answer using residue theorem.

[Ende Jin]

Why "greater than $2 \pi$" does not matter?
I see that you didn't emphasize $\beta$ to be a natural number, which is a bit weird because I don't even know how many singularities there are.

[Victor Ivrii]

Why "greater than $2 \pi$" does not matter?
I see that you didn't emphasize $\beta$ to be a natural number, which is a bit weird because I don't even know how many singularities there are.

We can reduce any $\beta>0$ to $\beta=2$ by substitution $x=t^{2/\beta}$; then $\gamma$ is replaced by $\delta=2(\gamma +1)/\beta -1$ and if $-1<\gamma<\beta-1$ (conditions needed to have convergency at $x=0$ and $x=\infty$ respectively), then $-1<\delta<1$.

We need only $\beta >0$; there will be just one singularity in the sector $0<\arg (z)< 2\pi/\beta$; namely as $z=e^{i\pi/beta}$.

[terryzhang]

Hi, Professor
If we do not use the ready formula for (c), does it mean that we need to prove this formula?

[Victor Ivrii]

If we do not use the ready formula for (c), does it mean that we need to prove this formula?

Yes, either in the general case, or in the special case of the given problem.