Integration of $e^{|x|} $ in computation of expectation I have a sequence of iid random variables $(X_n,n\in\mathbb{N})$ with density $f(x)= \frac{1}{2} e^{-|x|}$.
I need to verify that $M_n = \exp(\theta\sum_{i=1}^n X_i - nA(\theta))$ is integrable, for some values of $\theta$. Now, I should have that: $$ E(|M_n|) = \prod_{i=1}^nE( |\exp(\theta X_i - n A(\theta))|) =\prod_{i=1}^nE( \exp(\theta X_i - n A(\theta))) $$
while computing the integral, I splitted it to get rid of the absolute value:
$$ \prod_{i=1}^n 1/2\bigg(\int_{-\infty}^0 \exp(x)\exp(\theta x-nA(\theta))dx +\int_{0}^{+\infty} \exp(-x)\exp(\theta x-nA(\theta))dx\bigg) $$
But then in the solution I found :
$$ \prod_{i=1}^n 1/2\bigg(\int_{0}^{\infty} \exp(x(\theta-1)-A(\theta))dx +\int_{0}^{\infty} \exp(-x(\theta+1)-A(\theta))dx\bigg) $$
I understand that $M_n\geq 0$ so that might be a reason to do this, but then I don't see how the two expressions might be equivalent, did I do something wrong? Can you explain to me what is the rationale behind the second expression?

Note that the last integral in your expression is equal to their first integral.
Moreover, if you substitute $x=-y$ in you second integral you will get their first integral. Note that when changing the boundaries of integration gives a minus sign as well as $dy = -d x$ which thus cancel each other out.
I hope this helps. If you have any comments or want to see the details, feel free to comment!
Edit The rationale behind this substitution is no more than just wanting something positive to integrate over/it looks better.
