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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.

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1 Answer 1

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Hint:

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

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  • $\begingroup$ Oh ok I got it. I knew that I just didn't do a change of variables: Separate into two integrals with 0 to $\infty$ and $-\infty$ limits Change variables ($x$ to $-x$) and order of limits in second integral Group into one integral and factor out a 2 $\endgroup$
    – Alex
    Aug 19, 2018 at 23:34

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