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What is the Fourier Transform of $\rect(2Bt)\cos[{\omega}_Ct + k_fm(t_k)t] $?

I got the following as the solution:

$$\DeclareMathOperator{\sinc}{sinc} \frac{1}{2} \frac{1}{2B} \sinc(\frac{\omega+{\omega}_C+k_fm(t_k) +}{4B}) + \frac{1}{2} \frac{1}{2B} \sinc(\frac{\omega-{\omega}_C-k_fm(t_k) +}{4B})$$

However, in the book it is given as:

$$ \frac{1}{2} \sinc(\frac{\omega+{\omega}_C+k_fm(t_k) +}{4B}) + \frac{1}{2} \sinc(\frac{\omega-{\omega}_C-k_fm(t_k) +}{4B})$$

Wolfram alpha shows this:

https://www.wolframalpha.com/input/?i=Fourier+transform+calculator&assumption=%7B%22F%22%2C+%22FourierTransformCalculator%22%2C+%22transformfunction%22%7D+-%3E%22+rect%282Bt%29cos%28pt%29%22&assumption=%7B%22F%22%2C+%22FourierTransformCalculator%22%2C+%22variable1%22%7D+-%3E%22t%22&assumption=%7B%22F%22%2C+%22FourierTransformCalculator%22%2C+%22variable2%22%7D+-%3E%22w%22

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    $\begingroup$ you should have $2B$ outside of the $\sinc$, so I'm pretty sure your book can't be right, unless $B$ is defined somewhere else. $\endgroup$
    – Marcus Müller
    Apr 19 '20 at 17:50
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If you define the $\textrm{rect}()$ function according to this definition, then your result is correct. However, there might be a problem with the definition of the $\textrm{rect}()$ function. If in your book they define that function to have unit area, then the result of the book would be correct.

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  • $\begingroup$ I have used this text of B.P Lathi :academia.edu/36135973/… $\endgroup$
    – Anwesa Roy
    Apr 20 '20 at 12:28
  • $\begingroup$ Can it be possible that the book has defined the rect function to have unit area? Please refer to page no. 212. Thank you. $\endgroup$
    – Anwesa Roy
    Apr 20 '20 at 12:30
  • $\begingroup$ @AnwesaRoy: I actually don't think so, but just check if the author defines rect() somewhere in the book. But it's actually not a problem, just make sure that you understand the correctness of the result, whatever the result in the book may be. Even the best books have some errors. $\endgroup$
    – Matt L.
    Apr 20 '20 at 13:14
  • $\begingroup$ They have defined the rect(t) function just the way you have defined. Thanks for helping me out :) $\endgroup$
    – Anwesa Roy
    Apr 20 '20 at 13:34

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