maths integration help please...

Question

₀∫ ^∏  [sin (n + (1/2) ) X] / sin X dx equals :

• (2n +1 ) x ∏ / 2

• 0

• ∏

• n∏

Answer 1

You could check the great book by Lighthill on An Introduction to Fourier Analysis and Generalised Functions.

 

There is a problem like this in that book. If one uses the theory of generalized functions, in particular 1 (I think)

which is a good function, and you use theorems about good functions, it is a trivial problem.

Answer 2

The integral does not converge, so none of the multiple choice answers fit. Take for example n = 1. Your integral is then

 

∫[0,pi] sin(3x/2)/sin(x) dx   (1)

 

By the angle addition rule for sine,

 

sin(3x/2) = sin(x)cos(x/2) + cos(x)sin(x/2)   (2)

 

Dividing (2) by sin(x) gives

 

sin(3x/2)/sin(x) = cos(x/2) + cos(x)sin(x/2)/sin(x)   (3)

 

By the double-angle formula for sine, sin(x) = 2sin(x/2)cos(x/2). So we can write

 

cos(x)sin(x/2)/sin(x) = cos(x)/(2cos(x/2))   (4)

 

Using the double-angle identity cos(x) = 2cos^2(x/2) - 1, we have

 

cos(x)/(2cos(x/2) = [2cos^2(x/2) - 1]/(2cos(x/2)) = cos(x/2) - 0.5sec(x/2)   (5)

 

By (3), (4), and (5),

 

sin(3x/2)/sin(x) = 2cos(x/2) - 0.5sec(x/2)

 

and so

 

∫[0,pi] sin(3x/2)/sin(x) dx = ∫[0,pi] 2cos(x/2) dx - ∫[0,pi] 0.5sec(x/2) dx   (6)

 

On the right-hand side of equation (6), the first integral converges whereas the second integral diverges. So (1) must diverge.

 

I suspect that there is a misprint, in which the denominator is supposed to be sin(x/2). With that correction, you are integrating Dirichlet's kernel D_n(x) over [0,pi]. Since

 

D_n(x) = 1 + 2 ∑[k = 1, n] cos(kx)

 

and ∫[0,pi] cos(kx) dx = 0 for all k, then ∫[0,pi] D_n(x) dx is pi, which is the third answer choice.