How do I integrate sec(x) sin(nx) over a specific interval?

In summary: But it's still a bit weird that the textbook says to compute this when it's already given. That certainly makes a difference. It's not hard to show that the integral of sec x cos(nx) over a 2pi range is the negative of that of sec x cos((n-2)x), for n > 2. But it's still a bit weird that the textbook says to compute this when it's already given.
  • #1
mathskier
30
1
So I know that sec(x) has period 2Pi, and it's even so I don't need to figure out coefficients for bn.

Let's take the limits of the integral to go from -3/2 Pi to 1/2 Pi. How do I integrate sec(x) sin(nx) dx?! Am I on the right path?

PS: I know that this doesn't satisfy the Dirichlet Theorem, but the textbook I'm using still says to compute it.
 
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  • #2
Seems to me that integral is zero for n even and indeterminate for n odd.
 
  • #3
well since [itex]\sec(x) = 1/\cos(x) [/itex] then i think you can express the cosine functions, both on the top and on the bottom as exponentials with imaginary argument. i think you can get to an integral that looks like

[tex] c_n = \frac{1}{2 \pi} \int_{-\pi}^{+\pi} \frac{2}{e^{i (n+1) x}+e^{i (n-1) x}} \ dx [/tex]

i think you can find an antiderivative of that. might still come out as indeterminate. but that's the integral you have to solve.
 
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  • #4
haruspex said:
Seems to me that integral is zero for n even and indeterminate for n odd.
I evaluated the first few odd n integrals using Mathematica, and it looks like ##a_{2n+1} = (-1)^n 2\pi##.
 
  • #5
vela said:
I evaluated the first few odd n integrals using Mathematica, and it looks like ##a_{2n+1} = (-1)^n 2\pi##.
The reason I thought it indeterminate is that integrating over a 2π range effectively cancels a +∞ with a -∞. E.g. with n=1, it's the integral of tan, which is log cos. An interval that spans π/2 will appear to give a sensible answer, but in reality it's undefined.
 
  • #6
mathskier indicated the wrong integrand. It should be sec x cos(nx), which is the integrand I used.
 
  • #7
vela said:
mathskier indicated the wrong integrand. It should be sec x cos(nx), which is the integrand I used.
That certainly makes a difference. It's not hard to show that the integral of sec x cos(nx) over a 2pi range is the negative of that of sec x cos((n-2)x), for n > 2.
 

1. What is a Fourier Series of Secant?

A Fourier Series of Secant is a mathematical representation of the secant function using a combination of sine and cosine functions. It is a way of expressing the secant function as an infinite sum of periodic functions, and is used in many areas of mathematics, physics, and engineering.

2. How is a Fourier Series of Secant calculated?

The Fourier Series of Secant is calculated by using the Fourier series formula, which involves integrating the function over one period and then using coefficients to represent the individual sine and cosine terms in the series. The coefficients can be found using various methods, such as the Euler-Fourier formula or the method of least squares.

3. What is the significance of the Fourier Series of Secant?

The Fourier Series of Secant has many applications in mathematics, physics, and engineering. It allows for the approximation of complex functions using simpler trigonometric functions, making it easier to analyze and solve problems in these fields. It is also used in signal processing, where it can be used to filter and manipulate signals.

4. Can a Fourier Series of Secant represent any function?

No, a Fourier Series of Secant can only represent functions that are periodic and have a finite number of discontinuities within one period. It cannot represent functions with infinite discontinuities or functions that are not periodic.

5. Are there any limitations to the Fourier Series of Secant?

One limitation of the Fourier Series of Secant is that it can only approximate a function within its specified domain. It may not accurately represent the function outside of this domain. Additionally, the convergence of the series may be slow for certain functions, making it less practical for some applications.

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