Proving the equality of two integrals

[Hummingbird25]

Homework Statement

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 $$n \geq 0$$

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

 

[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
$$a_n = \int_0^{\pi} \frac{\sin x}{x - n\pi} \, dx$$
instead?

 

[Hummingbird25]

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
$$a_n = \int_0^{\pi} \frac{\sin x}{x - n\pi} \, dx$$
instead?

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 appreciate 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

 

[morphism]

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

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

 

[Hummingbird25]

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

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

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 $$\frac{dy}{dx} = 1$$

 

[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.

 

[Hummingbird25]

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.

Sorry then

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


I get the anti-derivative of the above to be

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

 

[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.

 

[Hummingbird25]

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.

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

 

[morphism]

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

 

[Hummingbird25]

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

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

Excuse me for asking but how does that simplify things?

 

[sutupidmath]

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

Excuse me for asking but how does that simplify things?

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

 

[Hummingbird25]

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

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

 

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

 

[Hummingbird25]

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

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

Hi and thank you for your answer,

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)

 

[e(ho0n3]

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

 

[sutupidmath]

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

Yep, and end of story!>.<

 

[Hummingbird25]

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

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.

 

[sutupidmath]

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.

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