- #1

- 302

- 0

[tex]

\int_0^{2\pi}d\phi f(a+b\cos\phi)\sin\phi=0

[/tex]

where a and b are arbitrary constants and f(x) is any function. Is this true? I guess it can be shown by expanding f in a power series of cosines?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- #1

- 302

- 0

[tex]

\int_0^{2\pi}d\phi f(a+b\cos\phi)\sin\phi=0

[/tex]

where a and b are arbitrary constants and f(x) is any function. Is this true? I guess it can be shown by expanding f in a power series of cosines?

- #2

Science Advisor

Homework Helper

- 25,836

- 256

hi daudaudaudau!

doesn't work for f = √, a = b

doesn't work for f = √, a = b

- #3

Gold Member

- 860

- 1

[tex]x=\cos{\phi}[/tex]

[tex]dx=-\sin{\phi}d\phi[/tex]

But this substitution is not 1-to-1: Each x value corresponds to 2 phi values on [0,2pi]. So you need to break the region of integration into [0,pi] and [pi,2pi]. If you look at the graph of the cos function, you will see it is 1-to-1 on these two intervals and goes from 1 to -1 on [0,pi] and from -1 to 1 on [pi,2pi]. So the integral becomes:

[tex]

\int_0^{2\pi}d\phi f(a+b\cos\phi)\sin\phi= \int_1^{-1}-f(a+bx)dx + \int_{-1}^{1}-f(a+bx)dx = \int_{-1}^{1}f(a+bx)dx + \int_{-1}^{1}-f(a+bx)dx = 0 [/tex]

Tiny-tim: I get this even in the example you gave.

- #4

Science Advisor

Homework Helper

- 25,836

- 256

yes, we can get it directly from the original integral …

if f = g', then it's ∫ [g(a + bcosφ)]' dφ, = [g(a + bcosφ)]

- #5

Gold Member

- 860

- 1

yes, we can get it directly from the original integral …

if f = g', then it's ∫ [g(a + bcosφ)]' dφ, = [g(a + bcosφ)]_{φ=0}^{2π}

Happens to everyone

That is a better way of proving it.

- #6

- 302

- 0

yes, we can get it directly from the original integral …

if f = g', then it's ∫ [g(a + bcosφ)]' dφ, = [g(a + bcosφ)]_{φ=0}^{2π}

What happened to the sine function?

- #7

Science Advisor

Homework Helper

- 25,836

- 256

chain rule

- #8

- 302

- 0

clever :-)

- #9

Science Advisor

Homework Helper

- 7,026

- 298

And the integral from -pi to pi equals the integral from 0 to 2pi.

QED.

Share:

- Replies
- 3

- Views
- 807

- Replies
- 16

- Views
- 741

- Replies
- 1

- Views
- 608

- Replies
- 5

- Views
- 622

- Replies
- 3

- Views
- 774

- Replies
- 14

- Views
- 1K

- Replies
- 3

- Views
- 316

- Replies
- 10

- Views
- 455

- Replies
- 11

- Views
- 386

- Replies
- 3

- Views
- 870