convolution of signals I'm finding in trouble trying to resolve this exercise. I have to calculate the convolution of two signals:
$$y(t)=e^{-kt}u(t)*\frac{\sin\left(\frac{\pi t}{10}\right)}{(\pi t)} $$
where $u(t)$ is Heavside function
well I applied the formula that says that the convolution of this two signal is equal to
$$Y(f)=X(f)W(f)$$
where $X(f)$ is the fourier transform of the first signal and $W(f)$ is the fourier transform of second signal
well fourier transform of $e^{-kt}u(t)$ is $X(f)=\frac{1}{k+j2\pi f}$. I have to make second signal as equals as possible to $\operatorname{sinc}\left(\frac{\pi t}{10}\right)$ so I do this operation:
$\frac{\sin\left(\frac{\pi t}{10}\right)}{\left(\frac{\pi t}{10}\right)}{\left(\frac{1}{10}\right)}$. this is equal ${\left(\frac{1}{10}\right)}\operatorname{sinc}\left(\frac{\pi t}{10}\right)$
right or not?
Edit
If something is not clear please advice me
 A: I think it is better to leave the Fourier transform idea, and use brute force instead.


*

*Write down the integral.

*Use the definition of the Heaviside function.

*Use a change of variable.

*Use Euler formula for $\sin$.

*End up with exponential integrals. 
A: I think this question should be solved in the frequency domain. Otherwise it seems difficult (at least to me).
$$e^{-kt}u(t)\rightarrow^F\frac{1}{k+jw}$$ and $$\frac{w_b}{\pi}sinc(wbt)\rightarrow^F rect(\frac{w}{2w_b})$$ and this in your case $w_b=\frac{\pi}{10}$
As a result we have $$Y(w)=X(w)W(w)=\frac{1}{k+jw}rect(\frac{w}{2w_b})$$ where $rect(\frac{w}{2w_b})=1$ if $|w|<wb$, else $0$. Finally what we need to do is to calculate the inverse fourier transform that is $$y(t)=F^{-1}(Y(w))=\frac{1}{2\pi}\int_{-w_b}^{w_b}\frac{1}{k+jw}e^{jwt}dw.$$ I think this integral should not be a big trouble.
EDIT: We can make a change of variables as $z=k+jw$ and $dz=jdw$. From here we have
$$\frac{1}{2\pi}\int_{-w_b}^{w_b}\frac{1}{k+jw}e^{jwt}dw=\frac{-j}{2\pi}\int_{-w_b}^{w_b}\frac{1}{z}e^{(z-k)t}dz=\frac{-j}{2\pi}\int_{-w_b}^{w_b}\frac{1}{z}e^{zt}dz\cdot\int_{-w_b}^{w_b}\frac{1}{z}e^{-kt}dz$$ from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions $$\int\frac{ e^{tz}}{z}dz=\ln|z|+\sum_{n=1}^{\infty}\frac{(tz)^n}{nn!}$$ and $$\int\frac{1}{z}e^{-kt}dz=\ln|z|e^{-kt}$$ Accordingly we have $$\frac{-j}{2\pi}\left[\ln|z|e^{-kt}\left(\ln|z|+\sum_{n=1}^{\infty}\frac{(tz)^n}{nn!}\right)\right]_{-w_b}^{w_b}$$ which is $$\frac{-j}{2\pi}\left(\ln w_be^{-kt}\left(\sum_{n=1}^{\infty}\frac{(tw_b)^n}{nn!}-\sum_{n=1}^{\infty}\frac{(-tw_b)^n}{nn!}\right)\right)$$
A: I think William DeMeo's comment is a good answer. You are simply trying to use the table and get your function into a form that can be used. 
The second function--call it $f_2$-- is currently $\frac{\sin (\pi t/10)}{\pi t}.$ Multiply it by 10, so you have $$f_3 = 10 \cdot f_2 = 10 \frac{\sin(\pi t/10)}{\pi t}= \frac{\sin(\pi t/10)}{(1/10)\pi t}= \text{sinc}(\pi t/10).$$  
Letting $f_1 = e^{-kt}u(t),$ we know that the product of the transforms $F_1$ and $F_3$ pairs with the convolution of the inverses of the functions in the time domain. 
So find the Fourier transforms of $f_1,f_3$, multiply those, and find the inverse transform of the product. That will be the convolution $c_1 =f_1*f_3.$ Since $AF(\omega)\leftrightarrow Af(t), c_2 = \frac{c_1}{10}$ is what you want. 
