Help needed regarding proof of definite integral problem

[rudra]

Homework Statement

f(x) is a bounded function and integrable on [a,b] . a, b are real constants. We have to prove that

i) A_n = _a∫^b f(x)cos(nx) dx → 0 when n→∞
ii)B_n = _a∫^b f(x)sin(nx) dx → 0 when n→∞

Homework Equations

Parseval's formula : For uniform convergence of f(x) with its Fourier series in [a,b]

_a∫^bf²(x) dx = L * ( a₀²/2 + Ʃ( a_n²+b_n² ) )


where a₀ a_n b_n are Fourier Coefficient. L = (b-a)/2

The Attempt at a Solution

If f(x) is bounded integrable in [-π,π]
Then by Parseval's Theorem
_-π∫^πf²(x) dx = L * ( A₀²/2 + Ʃ( A_n²+B_n² ) )

Hence Ʃ( A_n²+B_n² ) ) ≤ 1/L * ( _-π∫^πf²(x) dx )

Hence Ʃ( A_n²+B_n² ) ) converges to a finite value.

Hence A_n → 0 when n→∞ and B_n → 0 when n→∞

The above proof is assumed the interval is [-π,π] .

But the problem is when interval is [a,b]. Fourier coefficient A_n is given by
(1/L)*_a∫^b f(x)cos(nπx/L) dx

Any idea How to proceed further?

 

[Ray Vickson]

Homework Statement

f(x) is a bounded function and integrable on [a,b] . a, b are real constants. We have to prove that

i) A_n = _a∫^b f(x)cos(nx) dx → 0 when n→∞
ii)B_n = _a∫^b f(x)sin(nx) dx → 0 when n→∞

Homework Equations

Parseval's formula : For uniform convergence of f(x) with its Fourier series in [a,b]

_a∫^bf²(x) dx = L * ( a₀²/2 + Ʃ( a_n²+b_n² ) )


where a₀ a_n b_n are Fourier Coefficient. L = (b-a)/2

The Attempt at a Solution

If f(x) is bounded integrable in [-π,π]
Then by Parseval's Theorem
_-π∫^πf²(x) dx = L * ( A₀²/2 + Ʃ( A_n²+B_n² ) )

Hence Ʃ( A_n²+B_n² ) ) ≤ 1/L * ( _-π∫^πf²(x) dx )

Hence Ʃ( A_n²+B_n² ) ) converges to a finite value.

Hence A_n → 0 when n→∞ and B_n → 0 when n→∞

The above proof is assumed the interval is [-π,π] .

But the problem is when interval is [a,b]. Fourier coefficient A_n is given by
(1/L)*_a∫^b f(x)cos(nπx/L) dx

Any idea How to proceed further?

Just apply a scale-location transformation of the form x = A*u + B, to transform the x-interval [a,b] into the u-interval [-π,π].