Contour integration (related to deformation of path)

In summary, using the principle of deformation of path, we can deduce the integrals of the form \int_0^\infty t^n \textbf{cos}(bt) e^{-at}dt and \int_0^\infty t^n \textbf{sin}(bt) e^{-at}dt by considering the integral of 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. By using Cauchy
  • #1
equalP
21
0

Homework Statement


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

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

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 [itex]\int_0^\infty x^n e^{-x} dx=n![/itex]

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

But I don't know how to solve (ii) and also (iii) by deformation of path...
(ii)
[itex]\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[/itex]
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.
 
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  • #2
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:

[tex]\mathop\int\limits_{\text{red}}= \mathop\int\limits_{\text{blue}}+\mathop\int\limits_{\text{green}}[/tex]

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,

[tex]\mathop\int\limits_{\text{red}}+ \mathop\int\limits_{\text{blue}}+\mathop\int\limits_{\text{green}}=0[/tex]

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.
 

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  • #3
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?
 
Last edited:
  • #4
equalP said:
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.
 
  • #5
I found I typed something wrong (the inequality) and it should be the following:
(a) Consider the integral of [itex]f(z)=z^n e^{-z}[/itex] along three directed smooth curves:
(i)[itex]\textbf{Im} z=0[/itex], (ii)[itex]z=Re^{i\theta}[/itex], where [itex]0\leq\theta\leq\phi[/itex], (iii)[itex]z=ce^{i\phi}t[/itex], where [itex]t[/itex] goes from [itex]0[/itex] to [itex]\infty[/itex]
(b) Find the bound for modulus of the integral on (ii).
Use the inequality [itex]\textbf{cos}\theta\geq1-\frac{2\theta}{\pi}[/itex] for [itex]0\leq\theta <\frac{\pi}{2}[/itex]
After plugging in and having [itex]\int_{\text{Im}z=0}z^ne^{-z}dz=2n![/itex]
I got
[itex]\int_{z=Re^{i\theta}}z^n e^{-z}dz=\int_{ce^{i\phi}t}z^n e^{-z}dz-2n![/itex] with [itex]R\rightarrow \infty[/itex]
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 [itex]\int_{z=Re^{i\theta}}z^n e^{-z}dz=-n![/itex] should be true in order to deduce the two intregrals
But I do not know how to get this result...
 
  • #6
equalP;4543493 It also gives the hint to solve this problem. (a) Consider the integral of [itex said:
f(z)=z^n e^{-z}[/itex] along three directed smooth curves:
(i)[itex]\textbf{Im} z=0[/itex], (ii)[itex]z=Re^{i\theta}[/itex], where [itex]0\leq\theta\leq\phi[/itex], (iii)[itex]z=ce^{i\phi}t[/itex], where [itex]t[/itex] goes from [itex]0[/itex] to [itex]\infty[/itex]
(b) Find the bound for modulus of the integral on (ii).
Use the inequality [itex]\textbf{cos}\theta\leq1-\frac{2\theta}{\pi}[/itex] for [itex]0\leq\theta <\frac{\pi}{2}[/itex]

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 [itex]\cos(t)[/itex] and [itex]1-2t/\pi[/itex] 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:

[tex]\int_0^{t} z^n e^{-z}dz[/tex]

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

[tex]|(Re^{it})^n|\leq R^n[/tex]

How about the second one? Well,

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

and therefore:

[tex]|e^{-Re^{it}}|\leq |e^{-R\cos(t)}|[/tex]

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

[tex]|e^{-R\cos(t)}|[/tex]

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

[tex]\lim_{R\to\infty}\int_0^t z^n e^{-z} dt,\quad z=Re^{it}, 0\leq t\leq \pi/2[/tex]
 
  • #7
I want to ask a few questions.
Why [itex]|(Re^{it})^n|\leq R^n[/itex], [itex]|e^{-Re^{it}}|\leq |e^{-R\cos(t)}|[/itex] are ≤ but not =?
And I do like this:
[itex]|e^{-R\text{cos}\theta}|\leq|e^{-R(1-\frac{2\theta}{\pi})}|[/itex]
[tex]\lim_{R\to\infty}|\int_{z=Re^{i\theta}} z^n e^{-z} dz|,\quad 0\leq \theta \leq \phi\\
=\lim_{R\to\infty}|\int_0^\phi (Re^{i\theta})^n e^{-(Re^{i\theta})} Rie^{i\theta}d\theta|\\
\leq \lim_{R\to\infty}|\int_0^\phi R^n e^{-R(1-\frac{2\theta}{\pi})} Rie^{i\theta}d\theta|\\
=\lim_{R\to\infty}|iR^{n+1}\int_0^\phi e^{-R(1-\frac{2\theta}{\pi})} e^{i\theta}d\theta|\\
\leq \lim_{R\to\infty}|\frac{iR^{n+1}}{e^R} \int_0^\phi e^{\frac{2\theta}{\pi}}d\theta|\\
=\lim_{R\to\infty}|\frac{iR^{n+1}}{e^R}\frac{\pi}{\theta} [e^{\frac{2\theta}{\pi}}]_0^\phi|\\
=\lim_{R\to\infty}|\frac{iR^{n+1}}{e^R}\frac{\pi}{\theta} [e^{\frac{2\theta}{\pi}}]_0^\phi|\\
=0\quad \because \lim_{R\to\infty}\frac{iR^{n+1}}{e^R}=0
[/tex]
Is it right?
 
  • #8
No. You have:

[tex]\int_0^{\phi} e^{-R(1-2t/\pi)}dt,\quad 0<\phi<\pi/2[/tex]

What's that? Then, if [itex]0<\phi<\pi/2[/itex], what's

[tex]\lim_{R\to\infty}\left\{\int_0^{\phi} e^{-R(1-2t/\pi)}dt\right\}[/tex]
 
  • #9
[tex]\int_0^{\phi} e^{-R(1-2t/\pi)}dt,\quad 0<\phi<\pi/2\\
=\frac{\pi}{2}e^{-R} [e^{\frac{2t}{\pi}}]_0^\phi|\\
=\frac{\pi}{2}e^{-R}(e^{\frac{2\phi}{\pi}}-1)\\
\leq\frac{\pi}{2}e^{-R}(e-1)
[/tex]
[tex]\lim_{R\to\infty}\left\{\int_0^{\phi} e^{-R(1-2t/\pi)}dt\right\}\\
\leq\lim_{R\to\infty}\left\{\frac{\pi}{2}e^{-R}(e-1)\right\}\\
=0
[/tex]
Is it right?
 
  • #10
That's still not right equalP. What is:

[tex]\int e^{-R} e^{(\frac{2R}{\pi})t} dt[/tex]
 
  • #11
O, I found my mistake...
I forgot the R...

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

1. What is contour integration?

Contour integration is a mathematical technique used to evaluate complex integrals by deforming the path of integration. It involves breaking down a complex contour into simpler contours and using the properties of analytic functions to evaluate the integral along each contour.

2. How is the path of integration deformed in contour integration?

The path of integration is deformed by using the properties of analytic functions, which are functions that are differentiable at every point in a given domain. The path is typically deformed to avoid singularities, branch cuts, or other obstacles that may make the integral difficult to evaluate.

3. What are the benefits of using contour integration?

Contour integration allows for the evaluation of complex integrals that may not be possible using traditional integration techniques. It also allows for the evaluation of integrals over infinite domains, and can simplify the calculation of integrals involving trigonometric and exponential functions.

4. What are some common applications of contour integration?

Contour integration has many applications in mathematics, physics, and engineering. It is commonly used in complex analysis, signal processing, and quantum mechanics. It is also used in the calculation of various physical quantities such as electric and magnetic fields, fluid flow, and heat transfer.

5. Are there any limitations to using contour integration?

One limitation of contour integration is that it can only be applied to integrals that can be expressed as complex integrals. It also requires a good understanding of complex analysis and the properties of analytic functions. Additionally, the choice of contour can greatly affect the accuracy and efficiency of the integration, so careful selection is necessary.

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