Limits of ${\int\limits_{-\infty}^\infty f(x)e^x\mathrm{d}x}$ changed to $(0,\infty)$ if $f(x)$ is even

Question

In a medical imaging paper the following integral is changed from: $$\int_{-\infty}^\infty f(x)e^x\mathrm{d}x$$ To: $$\int_0^\infty2f(x)\cosh(x)\,\mathrm{d}x$$

I understand how you could get $2f(x)$ without the exponential but I don't get how one can deal with the exponentials without assuming they are a constant wrt $x$ that can be moved outside the integral and back in afterwards.

Answer 1

Hint:

$\cosh(x)=\dfrac{e^x+e^{-x}}2$