Darian Hubbard

Answered

2022-07-21

I ran into some trouble understanding how my professor went from this step in his solution:
${\int }_{-\pi }^{\pi }{|\frac{1}{\sqrt{2\pi }}{e}^{imx}-\frac{1}{\sqrt{2\pi }}{e}^{inx}|}^{2}\mathrm{d}x$
${\int }_{-\pi }^{\pi }{|\frac{1}{\sqrt{2\pi }}{e}^{imx}-\frac{1}{\sqrt{2\pi }}{e}^{inx}|}^{2}\mathrm{d}x$
to this step:
${\int }_{-\pi }^{\pi }\frac{1}{2\pi }\left(1-{e}^{i\left(m-n\right)x}-{e}^{i\left(n-m\right)x}+1\right)\mathrm{d}x$
I cannot see how this was done, please help!

Answer & Explanation

kamphundg4

Expert

2022-07-22Added 20 answers

I would recommend using the complex conjugate to calculate the squared absolute value.
${\int }_{-\pi }^{\pi }{|\frac{1}{\sqrt{2\pi }}{e}^{imx}-\frac{1}{\sqrt{2\pi }}{e}^{inx}|}^{2}\mathrm{d}x$
${\int }_{-\pi }^{\pi }\left(\frac{1}{\sqrt{2\pi }}{e}^{imx}-\frac{1}{\sqrt{2\pi }}{e}^{inx}\right)\left(\frac{1}{\sqrt{2\pi }}{e}^{-imx}-\frac{1}{\sqrt{2\pi }}{e}^{-inx}\right)\mathrm{d}x$
${\int }_{-\pi }^{\pi }\left(\frac{1}{2\pi }-\frac{1}{2\pi }{e}^{i\left(n-m\right)x}-\frac{1}{2\pi }{e}^{i\left(m-n\right)x}+\frac{1}{2\pi }\right)\mathrm{d}x$
${\int }_{-\pi }^{\pi }\frac{1}{2\pi }\left(1-{e}^{i\left(m-n\right)x}-{e}^{i\left(n-m\right)x}+1\right)\mathrm{d}x$

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