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[tex]P(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2} x^2},[/tex]

then

[tex]I=\int^{+\infty}_{-\infty}dx \, P(x) \theta(x)

= \int^{+\infty}_0 dx \, P(x)=\frac{1}{2},[/tex]

where [tex]\theta[/tex] is the step function. Now, using its integral representation, we have

[tex]

I=\int^{+\infty}_{-\infty}dx \, P(x) \frac{1}{2\pi i} \int^{+\infty}_{-\infty} \frac{dk}{k} \, e^{ikx} = \frac{1}{2\pi i} \int^{+\infty}_{-\infty} \frac{dk}{k} \int^{+\infty}_{-\infty}dx \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2+ikx}= \frac{1}{2\pi i}\int^{+\infty}_{-\infty} \frac{dk}{k} e^{-\frac{1}{2} k^2} .

[/tex]

But the integral over [tex]k[/tex] is calculated using residues and gives [tex]2\pi i[/tex] times the residue at [tex]k=0[/tex], what gives simply [tex]2\pi i[/tex] for the integral over [tex]k[/tex] and the wrong result 1 for the integral [tex]I[/tex]. I cannot see the catching. What is wrong with this calculation?

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# A Little Problem with a Simple Integral

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