- #1

equalP

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## 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 learnt 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.