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Suppose that $X$ follows an Exponential Distribution, then by definition of its inherent memorylessness property:

$$ \Pr[X > x + c \mid X > c] = \Pr[X > x] $$

I am wondering how this result translates to:

$$ {\rm E}[X \mid X > c] = {\rm E}[X] + c $$

Where exactly does the $c$ come from if I were to integrate?

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    \begin{align}\mathsf E[X \mid X>c]&=\int_0^\infty \mathsf P[X>x \mid X>c] dx\\&=\int_0^c 1 \cdot dx+\int_0^\infty \mathsf P[X>x+c \mid X>c]dx\\&=c + \int_0^\infty \mathsf P[X>x]\\&=\mathsf E[X]+c\end{align} – 2017-01-09
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    And some herustic argument: $\mathsf P[X>x+c \mid X>c]=\mathsf P[X>x]$ means $X-c$ under condition $X>c$ has the same distribution as $X$ under no condition. So $\mathsf E[X-c \mid X>c]=\mathsf E[X]$. – 2017-01-09

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The exponential distribution is $$ e^{-x/\mu}\ \frac{dx} \mu \text{ on } [0,\infty), $$ from which it follows that $$ \operatorname{E}(X) = \mu. $$ What is the conditional distribution of $X$ given the event $X>c\,\text{?}$ We have $$ \int_c^\infty e^{-x/\mu}\ \frac{dx} \mu = \int_{c\mu}^\infty e^{-u}\,du = e^{-c/\mu}, $$ and so $$ \int_c^\infty e^{c/\mu} e^{-x/\mu} \ \frac{dx} \mu = 1. $$ Therefore the conditional distribution of $X$ given $X>c$ is $$ e^{-(x-c)/\mu}\, \frac{dx}\mu \text{ on } [c,\infty). $$ This is just the original distribution shifted to the right by $c$ units. Thus the expected value must also be shifted to the right by $c$ units.