Solving Fourier Integration: Find Fourier Expansion for f(t)

[Paddy]

I am studying Fourier for an exam and came across something in my notes that I can't get my head round, might be a simple integration issue. Let me explain.

Homework Statement

The tutorial question in my notes that I am studying is as following:

1. Consider the periodic function defined by f(t) = {$$\frac{-1 \ \ \ \ -\pi \leq t \leq 0}{1 \ \ \ \ 0 < t < \pi}$$
Find its Fourier expansion.

Homework Equations

a₀ = 0 (because odd function)
a_n = 0 (because odd function)
b_n = $$\frac{2}{\pi} \int^{\pi}_{0} f(t) \ sin \ nt \ dt$$[/color]

3. The Solution written on my notes:
b_n = $$\frac{2}{\pi} \int^{\pi}_{0} 1 \ sin \ nt \ dt$$
b_n = $$\frac{2}{\pi} \left[-\frac{1}{n} \ cos \ nt\right]^{\pi}_{0}$$


My question is, how can you get $$-\frac{1}{n} \ cos \ nt$$ when integrating $$1 \ sin \ nt$$.

Should it not have been $$t \ cos \ nt$$ if integrating with respect to t (dt)?

I know it might be a simple answer but I have been studying for a while now and can't get my head round this, are my notes incorrect?

Note: I have it worked out in my notes down to the solution where $$f(t) = \frac{4}{\pi}(sint+\frac{1}{3}sin3t+\frac{1}{5}sin5t+\frac{1}{7}sin7t+...)$$ I have omitted most of the working out and most of my notes as they are irrelevant to my question.

 

[Dick]

To integrate 1*sin(nt)=sin(nt) you just substitute u=n*t, du=n*dt. You don't get t*cos(nt). That's just wrong. Try differentiating t*cos(nt) (use the product and chain rules). You don't get sin(nt).

 

[HallsofIvy]

I am studying Fourier for an exam and came across something in my notes that I can't get my head round, might be a simple integration issue. Let me explain.

Homework Statement

The tutorial question in my notes that I am studying is as following:

1. Consider the periodic function defined by f(t) = {$$\frac{-1 \ \ \ \ -\pi \leq t \leq 0}{1 \ \ \ \ 0 < t < \pi}$$
Find its Fourier expansion.

Homework Equations

a₀ = 0 (because odd function)
a_n = 0 (because odd function)
b_n = $$\frac{2}{\pi} \int^{\pi}_{0} f(t) \ sin \ nt \ dt$$[/color]

3. The Solution written on my notes:
b_n = $$\frac{2}{\pi} \int^{\pi}_{0} 1 \ sin \ nt \ dt$$
b_n = $$\frac{2}{\pi} \left[-\frac{1}{n} \ cos \ nt\right]^{\pi}_{0}$$


My question is, how can you get $$-\frac{1}{n} \ cos \ nt$$ when integrating $$1 \ sin \ nt$$.

Should it not have been $$t \ cos \ nt$$ if integrating with respect to t (dt)?

Surely you know better than that! The integral of sin(nt) is (-1/n)cos(nt) because the derivative of cos(nt) is - n sin(nt). I have no idea why you would want to multiply by "t"!

I know it might be a simple answer but I have been studying for a while now and can't get my head round this, are my notes incorrect?

Note: I have it worked out in my notes down to the solution where $$f(t) = \frac{4}{\pi}(sint+\frac{1}{3}sin3t+\frac{1}{5}sin5t+\frac{1}{7}sin7t+...)$$ I have omitted most of the working out and most of my notes as they are irrelevant to my question.

 

[Paddy]

Right, I understand Fourier no problem, its the stupid simple integration that is giving me headaches :(

if it was $$\int 1 + sin(nt)$$ then you would get $$\left[t + \frac{1}{n} cos nt \right]$$ Wouldn't you?

 

[Dick]

You would get t-(1/n)*cos(nt). There's a sign problem. But there's no product rule for integration. Integrating 1*sin(nt) has nothing to do with integrating 1.

 

[Paddy]

Right, sorted cheers!