# Mathematica Integration Question

1. May 3, 2010

### frenzal_dude

Hi, I need to find this integral:

$$G(f)=-A\int_{\frac{-3T}{2}}^{\frac{-T}{2}}e^{-j2\pi ft}dt+A\int_{\frac{-T}{2}}^{\frac{T}{2}}e^{-j2\pi ft}dt-A\int_{\frac{T}{2}}^{\frac{3T}{2}}e^{-j2\pi ft}dt$$

Here's the working out I did:

$$G(f)=-A[\frac{e^{j\pi fT}-e^{j\pi f3T}}{-j2\pi f}]+\frac{A}{\pi f}[\frac{e^{j\pi fT}-e^{-j\pi fT}}{2j}]-A[\frac{e^{-j\pi f3T}-e^{-j\pi fT}}{-j2\pi f}]$$

Therefore:

$$G(f)=2TAsinc(fT) -3TAsinc(3fT)$$

But when I used Mathematica I typed this:
Integrate[-A*Exp[-I*2*\[Pi]*f*t], {t, -T/2, (-3*T)/2}] +
Integrate[A*Exp[-I*2*\[Pi]*f*t], {t, -T/2, T/2}] +
Integrate[-A*Exp[-I*2*\[Pi]*f*t], {t, T/2, (3*T)/2}]

and it gave me this:

$$\frac{ASin(f\pi T)}{f\pi}-\frac{Ae^{-2j\pi ft}Sin(f\pi T)}{f\pi}+\frac{Ae^{2j\pi ft}Sin(f\pi t)}{f\pi}$$

which equals:

ATsinc(fT) + 2jATsinc(πFT)Sin(2πfT)

Is the answer from Mathematica the correct one?

2. May 4, 2010

### jackmell

Looks like you switched the integration limits in the first integral. Also, don't use capital letters for user-variables and you can omit the asterisks for multiplication if you use a space and also you can use FullSimplify to simplify the answer. Finally, and this is just a personal opinion: if I do it manually and then set it up correctly in Mathematica and the integrals are well-behaved, and I get two different answers that I can't show it's really the same answer written differently, good chance I'm the one wrong but there are rare exceptions.

3. May 4, 2010

### frenzal_dude

ahh thakyou so much! I fixed up the integral and used FullSimplify and now I got the same answer as when I use the Fourier Transform properties, so I know that my working out for the integral must be wrong some where.
Thanks again!

btw why can't we use capital letters as variables in mathematica?

Last edited: May 4, 2010