a challenging integral (perhaps)

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March 12th 2010, 07:44 AM
Random Variable

a challenging integral (perhaps)

$\int^{\pi}_{-\pi} \frac{\sin nx}{(1+2^{x}) \sin x} \ dx$ for n =0,1,2,...
March 12th 2010, 08:12 AM
Krizalid

put $x\mapsto-x$ and your integral equals $\int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx},$ thus $\int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}\,+\,\int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}=\int_{-\pi }^{\pi }{\frac{\sin nx}{\sin x}\,dx}=2\int_{0}^{\pi }{\frac{\sin nx}{\sin x}\,dx}.$

on the last integral let $I_n=\int_0^\pi\frac{\sin nx}{\sin x}\,dx$ and get that $I_n-I_{n-2}=0,$ thus $I_n=\left\{\begin{array}{cl}\pi,&\text{if }n\text{ is odd.}\\<br /> 0,&\text{if }n\text{ is even.}\end{array}\right.$
March 12th 2010, 08:36 AM
Random Variable

Quote:

Originally Posted by Krizalid

put $x\mapsto-x$ and your integral equals $\int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx},$ thus $\int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}\,+\,\int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}=\int_{-\pi }^{\pi }{\frac{\sin nx}{\sin x}\,dx}=2\int_{0}^{\pi }{\frac{\sin nx}{\sin x}\,dx}.$

on the last integral let $I_n=\int_0^\pi\frac{\sin nx}{\sin x}\,dx$ and get that $I_n-I_{n-2}=0,$ thus $I_n=\left\{\begin{array}{cl}\pi,&\text{if }n\text{ is odd.}\\<br /> 0,&\text{if }n\text{ is even.}\end{array}\right.$

There shouldn't be a two in front of $I_{n}$

$\int^{\pi}_{-\pi} \frac{\sin nx}{(1+2^{x})\sin x} \ dx = \int_{-\pi}^{0} \frac{\sin nx}{(1+2^{x})\sin x} \ dx + \int^{\pi}_{0} \frac{\sin nx}{(1+2^{x})\sin x} \ dx$

$= \int_{0}^{\pi} \frac{\sin nx}{(1+2^{-x})\sin x} \ dx + \int^{\pi}_{0} \frac{\sin nx}{(1+2^{x})\sin x} \ dx$

$= \int_{0}^{\pi} \frac{2^{x} \sin nx}{(1+2^{x})\sin x} \ dx + \int^{\pi}_{0} \frac{\sin nx}{(1+2^{x})\sin x} \ dx$

$= \int_{0}^{\pi} \frac{\sin nx}{\sin x} \ dx$
March 12th 2010, 08:57 AM
Random Variable

and for anyone who cares

$I_{n}-I_{n-2} = \int^{\pi}_{0} \frac{\sin nx}{\sin x} \ dx - \int^{\pi}_{0} \frac{\sin (n-2)x}{\sin x} \ dx$

$= \int^{\pi}_{0} \frac{\sin nx - \sin (n-2)x}{\sin x} \ dx$

$= 2 \int^{\pi}_{0} \frac{\sin x \cos (n-1)x}{\sin x} \ dx = 2 \int^{\pi}_{0} \cos (n-1)x \ dx$

$= \frac{2}{n-1} \sin (n-1)x \Big|^{\pi}_{0} = 0$

and obviously $I_{0} = 0$

and $I_{1} = \int^{\pi}_{0} \frac{\sin x}{\sin x} \ dx = \int^{\pi}_{0} \ dx = \pi$
March 12th 2010, 09:00 AM
Krizalid

note that i never said that your integral equals $2I_n,$ the latter was because $\frac{\sin nx}{\sin x}$ is even, what i actually did was

$\int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}=\frac{1}{2}\left( \int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}+\int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx} \right),$ and that yields the same you got.
March 12th 2010, 09:04 AM
Random Variable

Quote:

Originally Posted by Krizalid

note that i never said that your integral equals $2I_n,$ the latter was because $\frac{\sin nx}{\sin x}$ is even, what i actually did was

$\int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}=\frac{1}{2}\left( \int_{-\pi }^{\pi }{\frac{2^{x}\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx}+\int_{-\pi }^{\pi }{\frac{\sin nx}{\left( 1+2^{x} \right)\sin x}\,dx} \right),$ and that yields the same you got.

OK. My bad.
March 12th 2010, 10:03 AM
PaulRS
A general formula can be found for:

Remember hence:

Note that: let ; then :

Thus: - that is to say multiplied by the coefficient of of that polynomial in there - (*)

(*) Because is for and 0 for all other integer value of .

In particular then:
- $I(2n+1, 1) = 2\pi$
- $I(n, 2) = 2 \pi \cdot n$ , since our answer is the number of pairs of integers (x, y) with $0\leq x,y\leq n-1$ such that $x+y = n -1$ , fix a valid $x$ and that determines $y$ .