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Lapsed engineer/scientist (now a patent practitioner) seeking to expand his math skills. Most of those skills lay in areas useful in optics such as sums & integrals (like the one below), complex analysis, differential equations, transforms, and data analysis.

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"Wir müssen wissen, wir werden wissen." - David Hilbert

The picture on the left represents the integration region for the following integral of functions of finite support:

$$T(\mathbf{u}',\mathbf{u}'') = \int_{\mathbb{R}^2} d^2 \mathbf{\sigma} \; S(\mathbf{\sigma}) P(\mathbf{u}'+\mathbf{\sigma}) P^*(\mathbf{u}''+\mathbf{\sigma})$$


10h
awarded  Enlightened
12h
comment A tough integral and its generalization:
We've seen the first integral sevreal times in the past week here on SE, e.g., here: math.stackexchange.com/questions/986377/… ... The second integral is undefined in general.
1d
revised Prove that $\int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2$
edited tags
1d
answered Prove that $\int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2$
1d
answered Integrating (floor(x) e^-x) from 0 to inf
1d
answered Does $\int_0^{\infty}\frac{x\hspace{1mm}dx}{x^3+1}$ converge?
1d
revised Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$
added 301 characters in body
2d
comment Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$
@Aram, anonymous: ignoring the pole at $z=1$ using a contour integration approach in this case may lead to an incorrect result. There is in fact a nonzero contribution from an infinitesimal bump under the positive axis in the neighborhood of $z=1$.
2d
answered how to integrate $\int_{0}^1 \sqrt{(e^x+e^{-x}+2)} dx $?
2d
revised Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$
added 1 character in body
2d
comment Inverse Laplace Transform with cosh -fick's law diffusion
You're going to have to figure out how to make that equation cleaer if you cannot use MathJax yet, because the paper costs US$36 to access.
2d
comment Inverse Laplace transform of $\frac{1}{s} \frac{\sqrt{s}-1}{\sqrt{s}+1}$
math.stackexchange.com/questions/982125/…
2d
revised Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$
added 4 characters in body
2d
answered Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$
Oct
23
comment $\int_{0}^\pi \frac{\sin(nx)}{\sin x} dx$
@Sampaio: thanks for catching that.
Oct
23
revised $\int_{0}^\pi \frac{\sin(nx)}{\sin x} dx$
deleted 1 character in body
Oct
22
answered $\int_{0}^\pi \frac{\sin(nx)}{\sin x} dx$
Oct
22
comment Writing the integral $ \int_{t}^{\infty}(\frac{1}{4\pi s^{3}})^{1/2}\frac{r(|x|-r)}{|x|}e^{-\frac{(|x|-r)^{2}}{4s}}ds$ in simpler form?
Better to consider a Taylor series from the integral form in my answer over $u$. There, the lowest order term is clearly $1/\sqrt{t}$ from integrating the Taylor series of the exponential.
Oct
22
comment Writing the integral $ \int_{t}^{\infty}(\frac{1}{4\pi s^{3}})^{1/2}\frac{r(|x|-r)}{|x|}e^{-\frac{(|x|-r)^{2}}{4s}}ds$ in simpler form?
@TKM: well, then, there you go. You can get a more precise approximation from the Taylor series of the erf.
Oct
22
comment Writing the integral $ \int_{t}^{\infty}(\frac{1}{4\pi s^{3}})^{1/2}\frac{r(|x|-r)}{|x|}e^{-\frac{(|x|-r)^{2}}{4s}}ds$ in simpler form?
For what? As $t \to \infty$? In that case, then the integral is simply $O(1/\sqrt{t})$ because, for small $x$, $\operatorname{erf}{(x)} = O(x)$.