I have a signal which is a combination of sine signal and dead signal (0) as defined below. It is periodic with period 1.
$$\begin{eqnarray}
x(t) &=& sin(2*pi*1000*t) \enspace for \enspace 0<t<1/10 \nonumber \\
&=& 0 \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace \enspace for \enspace1/10 < t < 1 \nonumber \\
\end{eqnarray}$$
Since it is periodic and continuous, Continuous Time Fourier Transform exists. How to find it? I tried the below way and the answer is too complex. Is there any other way or am I doing something wrong?
My Solution:
$$T=1$$
$$w_0=2*\pi/T = 2*\pi$$
$$X(k)=1/T * \int_0^T x(t)*exp(-jkw_0t) \, dt$$
On solving I got, $$X(k)=\frac{1}{4*\pi} * \left\{ exp(-j2*\pi*k/10) * \left( \frac{exp(j200*\pi)}{k-1000} - \frac{exp(-j200*\pi)}{k+1000} \right) - \frac{2000}{k^2-1000000} \right\} $$
Here is a photo for reference on how I arrived to this equation. 
Any help will be greatly appreciated. Thanks!
You answer looks correct, but it doesn't provide much for one's intuition to verify its correctness, unfortunately.
You have a "sine multiplied by a shifted rectangle", so your Fourier Transform/Series coefficients should intuitively be something like "a pair of delta functions convolved with a modulated sinc() function".
Starting from your intermediate step
$$\begin{align*}X(k) &= \dfrac{1}{2j}\left[\dfrac{e^{-j2\pi(k-1000)t}}{-j2\pi(k-1000)}-\dfrac{e^{-j2\pi(k+1000)t}}{-j2\pi(k+1000)}\right]^{\frac{1}{10}}_0 \\ \\ &=\dfrac{1}{2j}\left[\dfrac{e^{-j2\pi(k-1000)\frac{1}{10}}-1}{-j2\pi(k-1000)}-\dfrac{e^{-j2\pi(k+1000)\frac{1}{10}}-1}{-j2\pi(k+1000)}\right]\\ \\ &=\dfrac{1}{2j}\left[e^{-j\pi(k-1000)\frac{1}{10}}\cdot\dfrac{e^{-j\pi(k-1000)\frac{1}{10}}-e^{j\pi(k-1000)\frac{1}{10}}}{-j2\pi(k-1000)}\\ -e^{-j\pi(k+1000)\frac{1}{10}}\cdot\dfrac{e^{-j\pi(k+1000)\frac{1}{10}}-e^{j\pi(k+1000)\frac{1}{10}}}{-j2\pi(k+1000)}\right]\\ \\ &=\dfrac{1}{20j}\left[e^{-j\pi(k-1000)\frac{1}{10}}\cdot\dfrac{e^{j\pi(k-1000)\frac{1}{10}}-e^{-j\pi(k-1000)\frac{1}{10}}}{2j\pi(k-1000)\frac{1}{10}}\\ -e^{-j\pi(k+1000)\frac{1}{10}}\cdot\dfrac{e^{j\pi(k+1000)\frac{1}{10}}-e^{-j\pi(k+1000)\frac{1}{10}}}{2j\pi(k+1000)\frac{1}{10}}\right]\\ \\ &=\dfrac{-j}{20}\left(e^{-j\frac{\pi}{10}(k-1000)}\cdot\mathrm{sinc}\left[\dfrac{\pi}{10}(k-1000)\right]\\ -e^{-j\frac{\pi}{10}(k+1000)}\cdot\mathrm{sinc}\left[\dfrac{\pi}{10}(k+1000)\right]\right)\\ \\ &=\dfrac{1}{10}e^{-j\frac{\pi}{10}k}\mathrm{sinc}\left(\dfrac{\pi}{10}k\right) *\dfrac{j}{2}\left[\delta(k+1000) - \delta(k-1000)\right] \end{align*}$$
Where $*$ denotes convolution, and $\mathrm{sinc}()$ is the non-normalized sinc function. This is the form of the result that my intuition predicted.
By the way, with the proper selection of the origin and period, such as having the center of the pulse at 0 and taking the base period from $-\frac{1}{2}$ to $+\frac{1}{2}$, you can get rid of the $e^{-j\frac{\pi}{10}k}$ modulation term. That simplifies the expression slightly.