Convolution of Fourier Transform

I have trouble calculating an example from the attached document (class material).
From my calculation, the answer should be

X(jw) = AT/2 ( sinc(T(w-w0)/(2pi)) + sinc(T(w+w0)/(2pi)) ).

The notation w (omega) is used in my calculation, and the attached example gives an answer in the notation f.
Even though the above equation was translated to the notation f, the solution in the attached ex. is little bit different from mine.
Could you try to solve the attached example in the notation w, to check my answer? If mine is incorrect, please advise the reason.

Thanks.

by using fourier transform equation over the limits -T/2 to T/2, you ca get the following equation

=A/2 {{ [e((j2pi(f-f0)T/2)-e((-j2pi(f-f0)T/2)]/(j2pi(f-f0)} + { [e((j2pi(f+f0)T/2)-e((-j2pi(f-f0)T/2)]/(j2pi(f+f0)}}
2 value in 2pi and T/2 gets cancelled in the exponential term, then
=A/2 {{[sin pi(f-f0)T]xT/pi(f-f0)T} + {[sin pi(f+f0)T]xT/pi(f+f0)T}
by taking T as common in the numerator term,
=AT/2 {{[sin pi(f-f0)T]/pi(f-f0)T} + {[sin pi(f+f0)T]/pi(f-f0)T}}
this is ur text book answer, there might be some mistake in your text book answer. it is showing 2pi in the final result instead of pi which is impossible.
when comes go the choice of w, the answer will e solved as
from the first eq which i mentioned, we can write the equation in the form of w as
=A/2 {{ [e((j(w-w0)T/2)-e((-j(w-w0)T/2)] x2/(2j(w-w0)} + { [e(j(w+w0)T/2)-e((-j(w+w0)T/2)]x2/(2j(w+w0)}}
by taking 2 as common in the numerator, it will get cancelled with A/2, and multiply th numerator and denominator with T/2
=A {T/2{[sin(w-w0)T/2]/T/2(w-w0)}+ {[sin(w+w0)T/2]/T/2(w+w0)}}
= AT/2 {sinc (w-w0)T/2 + sinc (w+w0)T/2}
this is ur final answer in w

dillikumar406, Thanks a lot for deriving.

From your following derivation to convert sin to sinc function, I'm just wondering the reason to use un-normalized sinc function (which is sinc(x) = sin(x)/x), instead of using normalized sinc (=sin(x pi)/(x pi)).

=A {T/2{[sin(w-w0)T/2]/T/2(w-w0)}+ {[sin(w+w0)T/2]/T/2(w+w0)}}
= AT/2 {sinc (w-w0)T/2 + sinc (w+w0)T/2}

If we use normalized sinc, then the answer becomes

= AT/2 {sinc ((w-w0)T/(2*pi)) + sinc ((w+w0)T/(2*pi))}

which is identical to the equation in my question. My concern is my derivation might be wrong due to use of normalized sinc.