Help with an integration problem

[vix_cse]

hi:

I am faced with an integration problem and can't seem to get even Maple/Mathematica to solve it. Would really appreciate if somebody can help me here. An approximate solution will also help.

$$\int_{-R}^{R}{cos(2\pi\nu x}) e^{-2\mu{\sqrt{R^2 - x^2}}}dx$$

thanks a lot for your time.

 

[topsquark]

hi:

I am faced with an integration problem and can't seem to get even Maple/Mathematica to solve it. Would really appreciate if somebody can help me here. An approximate solution will also help.

$$\int_{-R}^{R}{cos(2\pi\nu x}) e^{-2\mu{\sqrt{R^2 - x^2}}}dx$$

thanks a lot for your time.

Just off the cuff I would recommend a complex integration approach. At least if R approaches infinity that ought to work nicely.

-Dan

 

[tacman]

Hi there,

I understand that you are struggling with an integration problem and are seeking help. Integrals can be tricky, so it's great that you are reaching out for assistance. I'm not familiar with Maple or Mathematica, but I will try to guide you through solving this problem.

First, let's break down the integral into smaller parts. We can start by rewriting the exponential term as e^{-2\mu\sqrt{R^2 - x^2}} = e^{-2\mu\sqrt{(R^2 - x^2)/R^2}}. Next, we can use the trigonometric identity cos(2\pi\nu x) = (e^{2\pi i\nu x} + e^{-2\pi i\nu x})/2.

Using these substitutions, our integral becomes:

\int_{-R}^{R}{\frac{e^{2\pi i\nu x}}{2} + \frac{e^{-2\pi i\nu x}}{2}} e^{-2\mu\sqrt{(R^2 - x^2)/R^2}}dx

Now, we can split this into two integrals and solve them separately. Let's start with the first one:

\int_{-R}^{R}{\frac{e^{2\pi i\nu x}}{2}e^{-2\mu\sqrt{(R^2 - x^2)/R^2}}}dx

We can use the substitution u = \sqrt{(R^2 - x^2)/R^2} to simplify this integral. This will give us du = -\frac{x}{R^2\sqrt{R^2 - x^2}}dx. Substituting this back into the integral, we get:

\int_{-1}^{1}{\frac{e^{2\pi i\nu R u}}{2}e^{-2\mu u}}du

This integral can be solved using integration by parts. Let's define f(u) = \frac{e^{-2\mu u}}{2} and g'(u) = e^{2\pi i\nu R u}. This gives us f'(u) = -\mu e^{-2\mu u} and g(u) = \frac{e^{2\pi i\nu R u}}{2\pi i\nu R}.