Proving the equality of two integrals

1. Feb 26, 2008

Hummingbird25

1. The problem statement, all variables and given/known data

Prove

$$\int_{n\pi}^{(n+1)\cdot \pi} \frac{sin(x)}{x} dx = (-1)^n \cdot a_n$$

where $$a_n = \int_{0}^{\pi} \frac{sin(x)}{n\cdot pi x} dx$$

and $$n \geq 0$$

3. The attempt at a solution

My Question is how I do prove that the left side equals the right side? Writting their respective Riemann sums?

Sincerely Yours
Hummingbird

2. Feb 26, 2008

morphism

Are you sure you wrote it out correctly? The second integrand isn't defined when n=0, and when n=1 they're certainly not equal.

Did you perhaps mean to write
$$a_n = \int_0^{\pi} \frac{\sin x}{x - n\pi} \, dx$$

3. Feb 26, 2008

Hummingbird25

Hi morphism,

I meant to write

$$a_n = \int_0^{\pi} \frac{\sin x}{x + n\pi} \, dx$$.

Do I write out each side as a series or sum maybe? I would much appriciate a hint if possible:)

Because

I $$\int_{n\pi}^{(n+1)\cdot \pi} \frac{sin(x)}{x} dx = sin(x)/x |_{n\pi}^{({n+1})\cdot \pi}$$

Sincerely Yours
Hummingbird

Last edited: Feb 26, 2008
4. Feb 26, 2008

morphism

Sorry - that's what I meant to write as well!

Try the substitution $y=x-n\pi$.

5. Feb 26, 2008

Hummingbird25

Just be clear you mean I need to find the n anti-derivate of the integral and prove that its the same as the integral on right?

If I do what you ask then $y=x+n\pi$ then I get that

$$\frac{1}{y} \cdot \int_{0}^{\pi} sin(x) dx$$ where $$\frac{dy}{dx} = 1$$

Last edited: Feb 26, 2008
6. Feb 26, 2008

morphism

I meant for the substitution to be used with the first integral you posted. Is that what you did? I don't understand how you got a \frac{1}{y} on the outside.

7. Feb 26, 2008

Hummingbird25

Sorry then

$$\int_{0}^{\pi} \frac{sin(x)}{y} dx$$

I get the anti-derivative of the above to be

$$\frac{-cos(x)}{n \dot \pi+x}$$

8. Feb 26, 2008

morphism

It seems like you aren't applying the substitution properly.

The integral in question is
$$\int_{n\pi}^{(n+1)\pi} \frac{\sin x}{x} \, dx.$$

If we let $y=x-n\pi$ then it becomes
$$\int_{0}^{\pi} \frac{\sin (y+n\pi)}{y+n\pi} \, dy.$$

(Notice the change of variables!)

Using a certain identity to simplify $\sin (y+n\pi)$ will wrap things up.

9. Feb 26, 2008

Hummingbird25

Excuse me for asking you mean the unit circle identity where the signe in front of sinus changes constantly as n increases?

10. Feb 26, 2008

morphism

I mean sin(x+y)=sin(x)cos(y)+sin(y)cos(x).

11. Feb 26, 2008

Hummingbird25

$$\int_0^{\pi} \frac{sin(n \cdot \pi) \cdot cos(y) + cos(n \pi) \cdot sin(y)}{y+n \pi}} dx$$

Excuse me for asking but how does that simplify things?

Last edited: Feb 26, 2008
12. Feb 26, 2008

sutupidmath

just so you know, you are integrating with respect to y now!

13. Feb 26, 2008

Hummingbird25

Thank You

$$\int_0^{\pi} \frac{sin(n \cdot \pi) \cdot cos(y) + cos(n \pi) \cdot sin(y)}{y+n \pi}} dy$$

Do I factor the above or what do I do ?

SR

HM

14. Feb 26, 2008

sutupidmath

well, i am not sure, but i think that factoring would really simplify things, because noteice that sin(npi), and cos(npi) are sonstants so you would end up with sth similar to
$$\int_0^{\pi} \frac{sin(n \cdot \pi) \cdot cos(y) + cos(n \pi) \cdot sin(y)}{y+n \pi}} dy=sin(n\cdot\pi)\int_0^{\pi}\frac{cos(y)}{y+n\pi}dy+cos(n\pi)\int_0^{\pi}\frac{sin(y)}{y+n\pi}dy= (-1)^{n} a_n$$

now $$a_n = \int_0^{\pi} \frac{\sin y}{y + n\pi} \, dy$$

Last edited: Feb 26, 2008
15. Feb 26, 2008

Hummingbird25

I can see that in order to achieve the result I need to get rid of this term here

$$sin(n\cdot\pi)\int_0^{\pi}\frac{cos(y)}{y+n\pi}dy$$

Changing this into a sum would that allow me to get rid of this?

It can't simple be that I set n = 0 zero in the first part of the integral to get rid of it?

Sincerely
Maria the (Hummingbird)

Last edited: Feb 26, 2008
16. Feb 26, 2008

e(ho0n3

If n is an integer, what can you say about the value of $\sin n\pi$? What about the value of $\cos n\pi$?

17. Feb 26, 2008

sutupidmath

Yep, and end of story!>.<

18. Feb 26, 2008

Hummingbird25

If n icreases $\sin n\pi$ always be zero while $\cos n\pi$ will have its sign in fron change from - to plus everytime n increases!

Sincerely Maria.

Last edited: Feb 26, 2008
19. Feb 26, 2008

sutupidmath

Yes $$sin(n\pi)=0,while---- cos(n\pi)=(-1)^{n}$$
You are done, have another look at post #14

Last edited: Feb 26, 2008