Proving the equality of two integrals

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In summary, the conversation discusses how to prove the equality of two integrals, one from nπ to (n+1)π and the other from 0 to π. The conversation suggests using a substitution and applying a certain identity to simplify the integral. It is also mentioned that n must be an integer for the proof to hold.
  • #1

Homework Statement



Prove

[tex]\int_{n\pi}^{(n+1)\cdot \pi} \frac{sin(x)}{x} dx = (-1)^n \cdot a_n[/tex]

where [tex]a_n = \int_{0}^{\pi} \frac{sin(x)}{n\cdot pi x} dx[/tex]

and [tex]n \geq 0[/tex]

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
 
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  • #2
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
[tex]a_n = \int_0^{\pi} \frac{\sin x}{x - n\pi} \, dx[/tex]
instead?
 
  • #3
morphism said:
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
[tex]a_n = \int_0^{\pi} \frac{\sin x}{x - n\pi} \, dx[/tex]
instead?

Hi morphism,

I meant to write

[tex]a_n = \int_0^{\pi} \frac{\sin x}{x + n\pi} \, dx[/tex].

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

Because

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

Sincerely Yours
Hummingbird
 
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  • #4
Sorry - that's what I meant to write as well!

Try the substitution [itex]y=x-n\pi[/itex].
 
  • #5
morphism said:
Sorry - that's what I meant to write as well!

Try the substitution [itex]y=x-n\pi[/itex].

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 [itex]y=x+n\pi[/itex] then I get that

[tex]\frac{1}{y} \cdot \int_{0}^{\pi} sin(x) dx[/tex] where [tex]\frac{dy}{dx} = 1[/tex]
 
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  • #6
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
morphism said:
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

[tex]\int_{0}^{\pi} \frac{sin(x)}{y} dx[/tex]


I get the anti-derivative of the above to be

[tex]\frac{-cos(x)}{n \dot \pi+x}[/tex]
 
  • #8
It seems like you aren't applying the substitution properly.

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

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

(Notice the change of variables!)

Using a certain identity to simplify [itex]\sin (y+n\pi)[/itex] will wrap things up.
 
  • #9
morphism said:
It seems like you aren't applying the substitution properly.

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

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

(Notice the change of variables!)

Using a certain identity to simplify [itex]\sin (y+n\pi)[/itex] 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?
 
  • #10
I mean sin(x+y)=sin(x)cos(y)+sin(y)cos(x).
 
  • #11
morphism said:
I mean sin(x+y)=sin(x)cos(y)+sin(y)cos(x).

[tex]\int_0^{\pi} \frac{sin(n \cdot \pi) \cdot cos(y) + cos(n \pi) \cdot sin(y)}{y+n \pi}} dx[/tex]

Excuse me for asking but how does that simplify things?
 
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  • #12
Hummingbird25 said:
[tex]\int_0^{\pi} \frac{sin(n \cdot \pi) \cdot cos(y) + cos(n \pi) \cdot sin(y)}{y+n \pi}} dx[/tex]

Excuse me for asking but how does that simplify things?

just so you know, you are integrating with respect to y now!
 
  • #13
sutupidmath said:
just so you know, you are integrating with respect to y now!

Thank You

[tex]\int_0^{\pi} \frac{sin(n \cdot \pi) \cdot cos(y) + cos(n \pi) \cdot sin(y)}{y+n \pi}} dy[/tex]

Do I factor the above or what do I do ?

SR

HM
 
  • #14
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
[tex]\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[/tex]

now [tex]a_n = \int_0^{\pi} \frac{\sin y}{y + n\pi} \, dy[/tex]
 
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  • #15
sutupidmath said:
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
[tex]\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[/tex]

now [tex]a_n = \int_0^{\pi} \frac{\sin y}{y + n\pi} \, dy[/tex]

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

[tex]sin(n\cdot\pi)\int_0^{\pi}\frac{cos(y)}{y+n\pi}dy[/tex]

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)
 
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  • #16
If n is an integer, what can you say about the value of [itex]\sin n\pi[/itex]? What about the value of [itex]\cos n\pi[/itex]?
 
  • #17
e(ho0n3 said:
If n is an integer, what can you say about the value of [itex]\sin n\pi[/itex]? What about the value of [itex]\cos n\pi[/itex]?
Yep, and end of story!>.<
 
  • #18
e(ho0n3 said:
If n is an integer, what can you say about the value of [itex]\sin n\pi[/itex]? What about the value of [itex]\cos n\pi[/itex]?

If n icreases [itex]\sin n\pi[/itex] always be zero while [itex]\cos n\pi[/itex] will have its sign in fron change from - to plus everytime n increases!Sincerely Maria.
 
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  • #19
Hummingbird25 said:
If n icreases [itex]\sin n\pi[/itex] always be zero while [itex]\cos n\pi[/itex] will have its sign in fron change from - to plus everytime n increases!

Sincerely Maria.

Yes [tex]sin(n\pi)=0,while----

cos(n\pi)=(-1)^{n}[/tex]
You are done, have another look at post #14
 
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What is the definition of an integral?

An integral is a mathematical concept that represents the area under a curve in a graph. It is used to calculate the total value of a function over a given interval.

How do you prove the equality of two integrals?

To prove the equality of two integrals, you must show that they have the same value by using mathematical properties such as the fundamental theorem of calculus or by manipulating the integrals algebraically.

What are some common properties used to prove the equality of integrals?

Some common properties used to prove the equality of integrals include the commutative property, the distributive property, and the substitution property. These properties allow us to manipulate the integrals and show that they have the same value.

Can two integrals with different functions be equal?

Yes, two integrals with different functions can be equal if they have the same area under the curve. This is because the value of an integral is not dependent on the specific function, but rather on the area under the curve.

Why is it important to prove the equality of two integrals?

Proving the equality of two integrals is important because it allows us to simplify complex integrals and make calculations easier. It also helps us to understand the relationship between different functions and how they can have the same value despite being different.

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