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the mean square The Mean Square
(with one standard deviation and several unusual ones)

aka Rob Johnson

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Feb
1
answered Proving the convergence of limit and series
Feb
1
awarded  Good Answer
Feb
1
revised Convergence of $\frac{\sqrt{a_{n}}}{n}$
add latex
Feb
1
answered $n$th derivative of $e^x \sin x$
Feb
1
comment Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
We restrict the domain of $\log(\cos(z))$ to be $\mathbb{C}\setminus\{x\in\mathbb{R}:|x|\ge\pi/2\}$
Feb
1
comment Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
With the branch cut that you describe from the book, the only part of the real line that is in the domain of $\log(\cos(z))$ is $\left(-\frac\pi2,\frac\pi2\right)$. However, it sounds like $\Xi$ is the boundary of the upper-half plane and we consider the limits of the values as they tend to $\Xi$. On $\Xi$, $\log(\cos(z))$ has imaginary part $-n\pi$ on $\left(\left(n-\frac12\right)\pi,\left(n+\frac12\right)\pi\right)$. However, off $\Xi$, the imaginary part of $\log(\cos(z))$ can take on any value. For "principal branch", are you thinking of the branches of $\arccos(x)$?.
Feb
1
comment Normal and standard distribution
As Michael Hardy says $z=\frac{t-\mu}{\sigma}$. If $t$ goes up to $x$ then $z$ would go up to $\frac{x-\mu}{\sigma}$.
Feb
1
comment Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
The domain is all of $\mathbb{C}$ except the part of $\mathbb{R}$ whose absolute value is greater than or equal to $\frac\pi2$. This means that the only part of the real line is $\left(-\frac\pi2,\frac\pi2\right)$.
Jan
31
comment Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
Suppose we take the contour $[-i,i]\cup[i,i+2\pi]\cup[i+2\pi,-i+2\pi]\cup[-i+2\pi,-i]$. This contour exists within your domain, but since it circles two poles clockwise, the values of $\log(\cos(z))$ would differ by $4\pi i$.
Jan
31
comment Euler-Maclarurin summation formula and regularization
I am not sure precisely what you mean. However, if we try to split up $$ \begin{align} \int_0^\infty x^a\,\mathrm{d}x &=\overbrace{\int_0^1x^a\,\mathrm{d}x}^{a\gt-1} +\overbrace{\int_1^\infty x^a\,\mathrm{d}x}^{a\lt-1}\\ &=\frac1{a+1}-\frac1{a+1} \end{align} $$ My guess is that you won't find a useful regularization of this integral.
Jan
31
comment Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
Yes. The residues never cancel.
Jan
31
comment Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
Ack! Trying to do too many things at once is not good. I think the answer is correct now. :-)
Jan
31
revised Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
fix it up so that it is correct, finally!
Jan
31
awarded  Nice Answer
Jan
31
comment Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
Indeed, there looks to be a sign problem. However, when I was talking about their results following, I was talking about the increase in exponent, in contrast to my previous error. I noticed the sign change after I made that comment, but before I added the line about the sign change.
Jan
31
revised Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
fix post ala Daniel Fischer
Jan
31
comment Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
Your cut on the intervals will not work. As Daniel Fischer pointed out in a comment to my now modified answer, the residue of $\tan(z)$ is $-1$ at each pole. They will never cancel. The author seems to be using the branch cut I suggested in my previous comment and their results follow.
Jan
31
revised Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
fix miscalculation
Jan
31
revised Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$
add diagram
Jan
31
answered Integrating $\int \frac{e^{ipx}}{(\cos x)^{a}} \frac{dx}{x- \xi}$