Evaluating $\int _{-1}^{e} \frac{1}{x}dx$ Here very easily by the Fundamental Theorem of Calculus
$$\int _{-1}^{e} \frac{1}{x}dx=\ln(e)-\ln(-1)$$ 
From Euler's identity $e^{i \pi}$=-1 we can easily deduce that $\ln(-1)=i \pi$. Thus the integral becomes 
$$\int _{-1}^{e} \frac{1}{x}dx=1-i \pi$$ But we know that calculating an integral is calculating area and area cannot be a complex number (Which is coming in this case) and moreover $\ln(0)=- \infty$ then how come the bounds $(-1,e)$ give a finite positive value? Whats wrong here?
 A: The function $1/x$ is not Riemann-integrable over $[-1,e]$ (just try plotting it and see what happens to the "area under the curve" near $x=0$, it has a non-integrable singularity there), which is why the integral you wrote down does not exist. You can't apply the fundamental theorem of calculus to such an undefined integral.
To actually make sense of the integral, you need to say that it is an improper integral, and say that you are looking for its Cauchy principal value. See this (using the same example) and this.
The reason you got a complex value where you expected a real value is that you incorrectly calculated the integral. The subintegrals over $[-1,0)$ and $(0,1]$ should cancel, and the correct (principal) value of the improper integral is
$$ \mathop{\mathrm{P.V.}}\int_{-1}^e \frac{dx}{x} = \int_1^e \frac{dx}{x} = 1. $$
See also this and this.
A: First of all, $\int\frac1xdx = \ln(x)$ only when $x > 0$. If you want to evaluate it for $x<0$, you should make a change of variable $x = -y$ and then apply it.
Another mistake is to think that $e^{i\pi} = -1$ ie $\ln(-1) = i\pi$. What about $e^{3i\pi} = -1$?
This exercise has nothing to do with complex analysis and here is the solution I would give :
\begin{eqnarray*}
\int_{-1}^e \frac{dx}x &=& \int_{-1}^0\frac{dx}x + \int_0^1\frac{dx}x + \int_1^e \frac{dx}x\\
&=& \int_1^0\frac{-dy}{-y} + \int_0^1\frac{dx}x + \int_1^e \frac{dx}x \qquad y = -x\\
&=& -\int_0^1\frac{dy}{y} + \int_0^1\frac{dx}x + \int_1^e \frac{dx}x\\
\end{eqnarray*}
Notice the first two terms cancelling out so you are left with :
$$\int_{-1}^e \frac{dx}x = \int_1^e \frac{dx}x = \ln(e) - \ln(1) = 1$$
