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

Paddy
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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)
an = 0 (because odd function)
bn = \frac{2}{\pi} \int^{\pi}_{0} f(t) \ sin \ nt \ dt[/color]

3. The Solution written on my notes:
bn = \frac{2}{\pi} \int^{\pi}_{0} 1 \ sin \ nt \ dt
bn = \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.
 
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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).
 
Paddy said:
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)
an = 0 (because odd function)
bn = \frac{2}{\pi} \int^{\pi}_{0} f(t) \ sin \ nt \ dt[/color]

3. The Solution written on my notes:
bn = \frac{2}{\pi} \int^{\pi}_{0} 1 \ sin \ nt \ dt
bn = \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.
 
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?
 
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.
 
Right, sorted cheers!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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