Contour integration (related to deformation of path)

[equalP]

Homework Statement

Use the principle of deformation of path to deduce
\int_0^\infty t^n \textbf{cos}(bt) e^{-at}dt=\frac{n!}{e^{n+1}}\textbf{cos}((n+1)\phi) and \int_0^\infty t^n \textbf{sin}(bt) e^{-at}dt=\frac{n!}{e^{n+1}}\textbf{sin}((n+1)\phi)
where a>0, b>0, c=\sqrt{a^2+b^2}, and \phi=\textbf{tan}^{-1}(\frac{b}{a}) for 0\leq\phi<\frac{\pi}{2}

It also gives the hint to solve this problem.
(a) Consider the integral of f(z)=z^n e^{-z} along three directed smooth curves:
(i)\textbf{Im} z=0, (ii)z=Re^{i\theta}, where 0\leq\theta\leq\phi, (iii)z=ce^{i\phi}t, where t goes from 0 to \infty
(b) Find the bound for modulus of the integral on (ii).
Use the inequality \textbf{cos}\theta\leq1-\frac{2\theta}{\pi} for 0\leq\theta <\frac{\pi}{2}

Homework Equations

Cauchy's Integral Formula and Cauchy's Integral Theorem (I have only learned these two in the topic of Contour integration)

The Attempt at a Solution

I know the question should be solved by comparing real and imaginary part, but I don't know how to evaluate the integrals and hence I follow the hint.
In the previous part, I have shown that \int_0^\infty x^n e^{-x} dx=n!

Hence, I can also solved the (a)(i) of the hint:
\int_{Im z=0} f(z)dz=2\int_0^\infty x^n e^{-x} dx=2n!

But I don't know how to solve (ii) and also (iii) by deformation of path...
(ii)
\int_{z=Re^{i\theta}}z^n e^{-z}dz=\int_0^\phi R^n e^{in\theta-Re^{i\theta}}Rie^{i\theta}d\theta \\=R^{n+1}i\int_0^\phi e^{i(n+1)\theta-Re^{i\theta}}d\theta
Then I do not know what should I do in the next step...
It has the similar case in (iii)

Can anyone help me? Thank you.

 

[jackmell]

How about I ask you this: See that (closed) contour down there? Now suppose I wish to integrate a function over that contour that is analytic inside and on the contour. Since the function is analytic, the integral from 1 to 10 over the red contour should be the same as if I start at the origin, integrate over the blue, then integrate down over the green to the point (10,0) right? Integrating from 0 to 10 is independent of how I get to the end point. So I could write:

\mathop\int\limits_{\text{red}}= \mathop\int\limits_{\text{blue}}+\mathop\int\limits_{\text{green}}

as long as you're careful to keep straight, the path directions of the integrals. However, if I just integrate over all of them in a counterclock-wise direction, then by Cauchy's Theoerm, since the function is analytic,

\mathop\int\limits_{\text{red}}+ \mathop\int\limits_{\text{blue}}+\mathop\int\limits_{\text{green}}=0

Well, that's the three integration paths you have in i, ii, and iii above. Now, I haven't worked it out but I'd start by just plugging in all those integrals into these two formulas and try to just muscle-through all the algebra to see what happens.

 

[equalP]

I get what you mean.
But the path in ii does not at infinity. Or I can just change the R to infinity by having limit?

 

[jackmell]

I get what you mean.
But the path in ii does not at infinity. Or I can just change the R to infinity by having limit?

Yes, we let R go to infinity.

 

[equalP]

I found I typed something wrong (the inequality) and it should be the following:

(a) Consider the integral of f(z)=z^n e^{-z} along three directed smooth curves:
(i)\textbf{Im} z=0, (ii)z=Re^{i\theta}, where 0\leq\theta\leq\phi, (iii)z=ce^{i\phi}t, where t goes from 0 to \infty
(b) Find the bound for modulus of the integral on (ii).
Use the inequality \textbf{cos}\theta\geq1-\frac{2\theta}{\pi} for 0\leq\theta <\frac{\pi}{2}

After plugging in and having \int_{\text{Im}z=0}z^ne^{-z}dz=2n!
I got
\int_{z=Re^{i\theta}}z^n e^{-z}dz=\int_{ce^{i\phi}t}z^n e^{-z}dz-2n! with R\rightarrow \infty
How can I use the inequality to get the bound for modulus of it?
Besides, I found the integral iii is used to deduce the two required integrals and \int_{z=Re^{i\theta}}z^n e^{-z}dz=-n! should be true in order to deduce the two intregrals
But I do not know how to get this result...

 

[jackmell]

f(z)=z^n e^{-z}[/itex] along three directed smooth curves:
(i)\textbf{Im} z=0, (ii)z=Re^{i\theta}, where 0\leq\theta\leq\phi, (iii)z=ce^{i\phi}t, where t goes from 0 to \infty
(b) Find the bound for modulus of the integral on (ii).
Use the inequality \textbf{cos}\theta\leq1-\frac{2\theta}{\pi} for 0\leq\theta <\frac{\pi}{2}

Ok, let's take it slow. First off, that bounds is a mistake and would lead us in the wrong direction. I mean just plot \cos(t) and 1-2t/\pi in the range from 0 to pi/2. What you get? I'll tell you. Cos(t) is always above that line except at the end points and that's good because we want to know the bounds for the integral:

\int_0^{t} z^n e^{-z}dz

over the circular arc z=Re^{it} for t in the range of 0 to pi/2. So plug in the substitution z->Re^{it} in that expression and compute an upper limit on the value of the integral as a function of R. The first term is easy:

|(Re^{it})^n|\leq R^n

How about the second one? Well,

e^{-z}=e^{-Re^{it}}=e^{-R\cos(t)}(\cos(R\sin(t)+i\sin(R\sin(t))

and therefore:

|e^{-Re^{it}}|\leq |e^{-R\cos(t)}|

Now, we have just shown by plotting the functions that \cos(t)\geq 1-2t/\pi in the interval [0,\pi/2]. Ok, if that is the case, then what can we say about the upper bound on the expression:

|e^{-R\cos(t)}|

in that interval and therefore, what can we say about the upper bound on the integral:

\lim_{R\to\infty}\int_0^t z^n e^{-z} dt,\quad z=Re^{it}, 0\leq t\leq \pi/2

 

[equalP]

I want to ask a few questions.
Why |(Re^{it})^n|\leq R^n, |e^{-Re^{it}}|\leq |e^{-R\cos(t)}| are ≤ but not =?
And I do like this:
|e^{-R\text{cos}\theta}|\leq|e^{-R(1-\frac{2\theta}{\pi})}|
\lim_{R\to\infty}|\int_{z=Re^{i\theta}} z^n e^{-z} dz|,\quad 0\leq \theta \leq \phi\\<br /> =\lim_{R\to\infty}|\int_0^\phi (Re^{i\theta})^n e^{-(Re^{i\theta})} Rie^{i\theta}d\theta|\\<br /> \leq \lim_{R\to\infty}|\int_0^\phi R^n e^{-R(1-\frac{2\theta}{\pi})} Rie^{i\theta}d\theta|\\<br /> =\lim_{R\to\infty}|iR^{n+1}\int_0^\phi e^{-R(1-\frac{2\theta}{\pi})} e^{i\theta}d\theta|\\<br /> \leq \lim_{R\to\infty}|\frac{iR^{n+1}}{e^R} \int_0^\phi e^{\frac{2\theta}{\pi}}d\theta|\\<br /> =\lim_{R\to\infty}|\frac{iR^{n+1}}{e^R}\frac{\pi}{\theta} [e^{\frac{2\theta}{\pi}}]_0^\phi|\\<br /> =\lim_{R\to\infty}|\frac{iR^{n+1}}{e^R}\frac{\pi}{\theta} [e^{\frac{2\theta}{\pi}}]_0^\phi|\\<br /> =0\quad \because \lim_{R\to\infty}\frac{iR^{n+1}}{e^R}=0<br />
Is it right?

 

[jackmell]

No. You have:

\int_0^{\phi} e^{-R(1-2t/\pi)}dt,\quad 0&lt;\phi&lt;\pi/2

What's that? Then, if 0&lt;\phi&lt;\pi/2, what's

\lim_{R\to\infty}\left\{\int_0^{\phi} e^{-R(1-2t/\pi)}dt\right\}

 

[equalP]

\int_0^{\phi} e^{-R(1-2t/\pi)}dt,\quad 0&lt;\phi&lt;\pi/2\\<br /> =\frac{\pi}{2}e^{-R} [e^{\frac{2t}{\pi}}]_0^\phi|\\<br /> =\frac{\pi}{2}e^{-R}(e^{\frac{2\phi}{\pi}}-1)\\<br /> \leq\frac{\pi}{2}e^{-R}(e-1)<br />
\lim_{R\to\infty}\left\{\int_0^{\phi} e^{-R(1-2t/\pi)}dt\right\}\\<br /> \leq\lim_{R\to\infty}\left\{\frac{\pi}{2}e^{-R}(e-1)\right\}\\<br /> =0<br />
Is it right?

 

[jackmell]

That's still not right equalP. What is:

\int e^{-R} e^{(\frac{2R}{\pi})t} dt

 

[equalP]

O, I found my mistake...
I forgot the R...

\int_0^{\phi} e^{-R(1-2t/\pi)}dt,\quad 0&lt;\phi&lt;\pi/2\\<br /> =\frac{\pi}{2R}e^{-R} [e^{\frac{2Rt}{\pi}}]_0^\phi\\<br /> =\frac{\pi}{2R}e^{-R}(e^{\frac{2R\phi}{\pi}}-1)\\<br /> \leq\frac{\pi}{2R}e^{-R}(e^R-1)\\<br /> =\frac{\pi}{2R}(1-e^{-R})<br />
\lim_{R\to\infty}\left\{\int_0^{\phi} e^{-R(1-2t/\pi)}dt\right\}\\<br /> \leq\lim_{R\to\infty}\left\{\frac{\pi}{2R}(1-e^{-R})\right\}\\<br /> =0-0=0<br />
Is it right?