A simpler method or more descriptive answer to the Fourier Transform

I'm trying to do a CT Fourier Transform of these two signals $$e^{-a(t-1)} \cdot u(t-1)$$ and $$e^{-a(t-1)} \cdot u(t)$$ Where $a$ is any real number, and $u(t)$ is the unit step function.

My question is if there is either a property that eliminates the need to evaluate the integral or if there is some simplification of the transform so it's not such a jumble.

I see that i made a mistake with integrating the unit step, and my answer should not have that extra $$t \cdot u(t-1)$$ and $$t \cdot u(t)$$ After the help i got on the DSP exchange and here i ended up with $$x_1(F)=\frac{e^{2a+j2\pi\cdot F}}{a+j2\pi\cdot F}$$ and $$x_2(F) =\frac{e^a}{a+j2\pi\cdot F}$$ That does seem more elegant than what i previously had, i feel like i could possibly manipulate $x_1(F)$ into a sinc function, but i don't think it's necessary.

• What is $e^($? Also, please rotate your image. – user17592 Apr 6 '13 at 19:26
• chattypics.com/files/droidUpload_qdg2kdu2di_yp4yyclp6d.jpg That is supposed to be e to the quantity -a(t-1), – retroredeye Apr 6 '13 at 19:32
• Okay.. I edited it in for you this time. What exactly is your question? – user17592 Apr 6 '13 at 19:35
• Thanks for fixing that up for me, i don't really get why that didn't work, but my question is if there is either a property that eliminates the need to evaluate the integral or if there is some simplification of the transform so it's not such a jumble. – retroredeye Apr 6 '13 at 19:40
• I edited the question in. You can edit your questions yourself as well. – user17592 Apr 6 '13 at 19:41

You have a definite integral with endpoints. Think about how the properties of $u(t)$ let you reduce the integration boundaries.
First, you should be able to find the transform for the time domain signal $f(t) = e^{-at}u(t)$ in your Fourier transform tables.
For your first signal, a table of Fourier transforms like this one will tell you that if the transform of $f(t)$ is $F(\omega)$, then the transform of $f(t-t_0)$ is $F(\omega)e^{-j\omega{}t_0}$.
For your second signal, realize that you can write $e^{-a(t-1)}$ as $e^{a}e^{-at}$, meaning this signal is just a constant multiplied by the basic decaying exponential signal.