### Author Topic: 2.6 Q22  (Read 3212 times)

#### Kathy Ngo

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##### 2.6 Q22
« on: November 22, 2018, 12:16:49 PM »
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

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##### Re: 2.6 Q22
« Reply #1 on: November 22, 2018, 02:12:09 PM »
Isn't it just substituting in the number?

#### Min Gyu Woo

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##### Re: 2.6 Q22
« Reply #2 on: November 22, 2018, 02:32:44 PM »
The steps of getting to that final equation. Not the formula.

#### ruienlin

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##### Re: 2.6 Q22
« Reply #3 on: November 22, 2018, 02:33:25 PM »
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

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##### Re: 2.6 Q22
« Reply #4 on: November 22, 2018, 02:37:01 PM »
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

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##### Re: 2.6 Q22
« Reply #5 on: November 22, 2018, 02:41:00 PM »
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

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##### Re: 2.6 Q22
« Reply #6 on: November 22, 2018, 03:59:18 PM »
You should not use ready formulae but to derive the answer using residue theorem.

#### Ende Jin

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##### Re: 2.6 Q22
« Reply #7 on: November 23, 2018, 07:25:37 PM »
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

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##### Re: 2.6 Q22
« Reply #8 on: November 24, 2018, 04:24:02 AM »
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 $z=e^{i\pi/beta}$.
« Last Edit: November 27, 2018, 01:53:01 PM by Victor Ivrii »

#### terryzhang

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##### Re: 2.6 Q22
« Reply #9 on: November 27, 2018, 11:59:41 AM »
Hi, Professor
If we do not use the ready formula for (c), does it mean that we need to prove this formula?
« Last Edit: November 27, 2018, 12:05:58 PM by terryzhang »