- #1
Susanne217
- 317
- 0
A never ending integral?(last post hopefully the solution
I am trying to solve the integral
[tex]\frac{1}{2\pi} \int_{0}^{\pi} e^{-inx} \cdot cos(x) dx [/tex]
but it freaking never ends :(
This should have been to be solve by integration by parts but I can never get the v in
[tex]uv - \int v du [/tex] to get small enough to end the integral
I take the orignal integral
[tex]\frac{1}{2\pi} \int_{0}^{\pi} e^{-inx} \cdot cos(x) dx [/tex]
let u = cos(x) and [tex]du = -sin(x) dx [/tex]
then [tex] v = \frac{i \cdot e^{-inx}}{n}[/tex] since [tex]dv = e^{-inx} dx[/tex]
thus [tex] \frac{1}{2\pi} \int_{0}^{\pi} e^{-inx} \cdot cos(x) dx = \frac{-sin(n\pi)}{n}+ (\frac{-cos(n\pi)}{n} - \frac{1}{n}) \cdot i - \int_{0}^{\pi} \frac{i \cdot e^{-inx}}{n} \cdot -sin(x) dx [/tex]
If I do integration by parts a second time I get
[tex]u = -sin(x), du = -cos(x) dx[/tex] and [tex]dv = \frac{i \cdot e^{-inx}}{n}[/tex] and thus [tex]v = \frac{i \cdot e^{-inx}}{n^2}[/tex]
Which gives me
[tex] -sin(x) \cdot \frac{i \cdot e^{-inx}}{n^2}|_{0}^{\pi} - \int_{0}_{\pi} \frac{i \cdot e^{-inx}}{n^2} \cdot -cos(x) dx [/tex]
and I then end up with an expression which looks like this
[tex]I = \frac{1}{2\pi} \int_{0}^{\pi} e^{-inx} \cdot cos(x) dx = \frac{-sin(n\pi)}{n}+ (\frac{-cos(n\pi)}{n} - \frac{1}{n}) \cdot i - \int_{0}^{\pi} \frac{i \cdot e^{-inx}}{n^2} \cdot -cos(x) dx [/tex]
Dick, by solving it you mean
[tex]I = \frac{1}{2\pi} \int_{0}^{\pi} e^{-inx} \cdot cos(x) dx + \int_{0}^{\pi} \frac{i \cdot e^{-inx}}{n^2} \cdot cos(x) dx = \frac{-sin(n\pi)}{n}+ (\frac{-cos(n\pi)}{n} - \frac{1}{n}) \cdot i [/tex]?
Sincerely
Susanne.
Homework Statement
I am trying to solve the integral
[tex]\frac{1}{2\pi} \int_{0}^{\pi} e^{-inx} \cdot cos(x) dx [/tex]
but it freaking never ends :(
Homework Equations
This should have been to be solve by integration by parts but I can never get the v in
[tex]uv - \int v du [/tex] to get small enough to end the integral
The Attempt at a Solution
I take the orignal integral
[tex]\frac{1}{2\pi} \int_{0}^{\pi} e^{-inx} \cdot cos(x) dx [/tex]
let u = cos(x) and [tex]du = -sin(x) dx [/tex]
then [tex] v = \frac{i \cdot e^{-inx}}{n}[/tex] since [tex]dv = e^{-inx} dx[/tex]
thus [tex] \frac{1}{2\pi} \int_{0}^{\pi} e^{-inx} \cdot cos(x) dx = \frac{-sin(n\pi)}{n}+ (\frac{-cos(n\pi)}{n} - \frac{1}{n}) \cdot i - \int_{0}^{\pi} \frac{i \cdot e^{-inx}}{n} \cdot -sin(x) dx [/tex]
If I do integration by parts a second time I get
[tex]u = -sin(x), du = -cos(x) dx[/tex] and [tex]dv = \frac{i \cdot e^{-inx}}{n}[/tex] and thus [tex]v = \frac{i \cdot e^{-inx}}{n^2}[/tex]
Which gives me
[tex] -sin(x) \cdot \frac{i \cdot e^{-inx}}{n^2}|_{0}^{\pi} - \int_{0}_{\pi} \frac{i \cdot e^{-inx}}{n^2} \cdot -cos(x) dx [/tex]
and I then end up with an expression which looks like this
[tex]I = \frac{1}{2\pi} \int_{0}^{\pi} e^{-inx} \cdot cos(x) dx = \frac{-sin(n\pi)}{n}+ (\frac{-cos(n\pi)}{n} - \frac{1}{n}) \cdot i - \int_{0}^{\pi} \frac{i \cdot e^{-inx}}{n^2} \cdot -cos(x) dx [/tex]
Dick, by solving it you mean
[tex]I = \frac{1}{2\pi} \int_{0}^{\pi} e^{-inx} \cdot cos(x) dx + \int_{0}^{\pi} \frac{i \cdot e^{-inx}}{n^2} \cdot cos(x) dx = \frac{-sin(n\pi)}{n}+ (\frac{-cos(n\pi)}{n} - \frac{1}{n}) \cdot i [/tex]?
Sincerely
Susanne.
Last edited: