# Hi! This is a test question. Please enjoy it reponsibly. Testing testing. fx==int_{infty}^{-infty} fxi​(e^2pixi x)dxi

Albarellak 2021-01-31 Answered
Hi! This is a test question. Please enjoy it reponsibly.
Testing testing.
$fx=={\int }_{\mathrm{\infty }}^{-\mathrm{\infty }}f\xi \left({e}^{2}\pi \xi x\right)d\xi$
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## Expert Answer

Elberte
Answered 2021-02-01 Author has 95 answers
Sup, bruh?
Let $a=b+1a=b+1.$ Then
$\left(a-b\right)a=\left(a-b\right)\left(b+1\right)$
${a}^{2}-ab=ab+a-{b}^{2}-{b}^{2}-b$
${a}^{2}-ab-a=ab+a-a-{b}^{2}-{b}^{2}-b$
$a\left(a-b-1\right)=b\left(a-b-1\right)$
$a=b$
$b+1=b$
Hence 1=0
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