I have to solve the following convolution:
$$y(t)=e^{-t}u(t)*\sum_{k=-\infty}^{\infty}\delta(t-2k)$$
Here $\delta(.)$ is Dirac delta function. The summation is:
$$\sum_{k=-\infty}^{\infty}\delta(t-2k)=\frac12\sum_{k=-\infty}^{\infty}\delta\left(k-\frac t2\right)=\frac12\operatorname{III}\left(\frac{t}{2}\right)$$
Here $\operatorname{III}(.)$ is the Dirac Comb function.
Now continuing with the definition of convolution, we have:
$$y(t)=\frac12\int_{-\infty}^{\infty}e^{-\tau}u(\tau)\sum_{k=-\infty}^{\infty}\delta\left(\left(k-\frac {t}{2}\right)-\tau\right)\ \mathrm{d}\tau$$
Using the "sifting" property of the Dirac Delta
$$\begin{align*}y(t)&=\frac12\sum_{k=-\infty}^{\infty}e^{-\left(k-\frac {t}{2}\right)}u\left(k-\frac {t}{2}\right)\\ &=\frac12e^{\frac {t}{2}}\sum_{k=\frac {t}{2}}^{\infty}e^{-k}\\ &=\frac12e^{\frac {t}{2}}\left(\frac{1}{e^{\frac {t}{2}}}+\frac{1}{e^{\frac {t}{2}+1}}+\frac{1}{e^{\frac {t}{2}+2}}+\cdot \cdot \cdot \right)\\ &=\frac12\left(1+\frac{1}{e^{1}}+\frac{1}{e^{2}}+\cdot \cdot \cdot \right)=\frac12\frac{1}{1-e^{-1}} \end{align*}$$
But the answer in my book is $\displaystyle \frac{1}{1-e^{-2}}e^{-t}$, $0\leq t<2$
Best Answer
There are actually two mistakes.
1) You get the convolution wrong. It should read $$ y(t)=\frac12\int_{-\infty}^{\infty}e^{-\tau}u(\tau)\sum_{k=-\infty}^{\infty}\delta\left(\left(k-\frac {t}{2}\right)+\frac12 \tau\right)\ \mathrm{d}\tau$$
2) There is a mistake in the line with $\sum_{k=t/2}^\infty$. Note that $k$ should be an integer!!!
Here, is a correct derivation. Let us start with the definition of the convolution $$ y(t)= \int e^{-\tau}u(\tau)\sum_{k=-\infty}^{\infty}\delta(t-2k-\tau) d\tau \,.$$
Then we use the sifting property to obtain $$ y(t)= \sum_{k=-\infty}^{\infty} e^{2k-t}u(t-2k)\;.$$
Now the summation over $k$ should include the integers that are smaller than $t/2$. In particular, for $0\leq t <2$, these include $k\in \{0,-1,-2,\dots\}$. We obtain $$ y(t)= \sum_{k=-\infty}^{0} e^{2k-t} = \frac{e^{-t}}{1-e^{-2}}$$ valid for $0\leq t <2$.