Expectation of Random Variable with even Probability Density Function By Definition of Expectation of Random Variable:
$$ E(X)= \int_{-\infty}^{\infty}\,x\,f_X(x)\,dx $$
Now if the pdf $f_X(x)$ is Even we know that $E(X)=0$ (Ofcourse if integral Converges, i.e, Lets exclude cases like Cauchy Random Variable)
Is the Converse True, i.e., is there a Random Variable $X$ whose pdf is Neither Even-Nor Odd, such that $E(X)=0$.
 A: It is possible to make up as many examples as you wish. Let $Y$ be almost any random variable with mean $\mu$, and let $X=Y-\mu$. All we need to do is to avoid symmetry about $\mu$, so for example let $Y$ have exponential distribution, or density $2y$ in $[0,1]$ and $0$ elsewhere.
For a discrete example, let $Y$ have binomial distribution with $p\ne 0$, $1$, or $1/2$. 
A: Define $f_X(x):=\frac 1{x^3}\chi_{[1,+\infty)}(x)+\frac 12\chi_{(-5/2,-3/2)}$. 


*

*Since $\int_{\Bbb R}f_X(x)dx=\left[-\frac 1{2x^2}\right]_1^{+\infty}+\frac 12=1$
and $f_X$ is non-negative, it's a density. 

*We have 
\begin{align}
E[X]&=\int_1^{+\infty}\frac 1{x^2}dx+\frac 12\int_{-5/2}^{-3/2}xdx\\\
&=\left[-\frac 1x\right]_1^{+\infty}+\frac 12\left[\frac{x^2}2\right]_{-5/2}^{-3/2}\\
&=1+\frac 14\frac 14(9-25)\\
&=0.
\end{align}

*$f_X$ is not even, since $f_X(10)=\frac 1{1000}\neq 0=f_X(-10)$ 


(there should be simpler examples)
Note that a density function cannot be odd, since in this case $\int_{-\infty}^{+\infty}f(x)dx=0$, whereas it should be $1$. 
