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 Apr13 comment Calculate $\int_\Gamma ze^{z}dz$ where $\Gamma$ is line from point $z_1=0$ to point $z_2=\frac{\pi i}{2}$ @user227317 yes. See my answer for full details, but you got the answer! You can also use Blatter's approach too. Apr12 comment Calculate $\int_\Gamma ze^{z}dz$ where $\Gamma$ is line from point $z_1=0$ to point $z_2=\frac{\pi i}{2}$ @user227317 no, as I said use the substitution $z=it$. Also @ JessicaK's solution will also work. See also @ ChrisrianBlatter's answer. Apr12 comment Calculate $\int_\Gamma ze^{z}dz$ where $\Gamma$ is line from point $z_1=0$ to point $z_2=\frac{\pi i}{2}$ You need to parametrize the curve $\Gamma$, e.g. let $z(t)=it$ where $t\in[0,\pi/2]$. Looks like integration by parts may be helpful too. Apr8 comment $F(x)+G(y)= e^{x+y}?$ Yes, the way it is written confused me for a moment. However, on expanding the middle equality now I clearly see it does equal $F(1)-F(0)$. I was looking at what you had written from a different perspective - I was trying to construct the middle equality from the first by rearranging the original equation. Sorted now I see clearly what's happing - that "old trick" of adding something and then taking it away, so yes maybe best read right to left. +1 for your answer! Apr8 comment $F(x)+G(y)= e^{x+y}?$ I could be wrong, but shouldn't the middle equality be $-G(y)+e^{1+y}-(-G(y)+e^y)$ ? Maybe it's equivalent... Apr8 comment How to deduce that $1\cdot 1 + 2\cdot 1 + 2\cdot 2 + 3\cdot 1+3\cdot 2+3\cdot 3 +…+(n\cdot n) = n(n+1)(n+2)(3n+1)/24$ Try proof by induction ? Mar30 comment Question on branches and $\iff$. Thank you. So if I restrict my attention to the principal branch (and I will also assume $f$ and $g$ are continuous) then $f+ig=0\iff e^{f(x)}\cos(g(x))+ie^{f(x)}\sin(g(x))=1$ ? Mar27 comment What does $O\left(\frac{1}{\log\log T}\right)$ mean? Thanks. That's what I thought. So basically what the paper says is that as $T\to\infty$ the number of zeros outside the region is some constant multiple of $1/(\log\log T)$ ? Mar25 comment Series with $e^{\frac{1}{n}}$ Shouldn't that be $+O(n^{-4})$ in your equality? Mar25 comment Series with $e^{\frac{1}{n}}$ You could first try expanding $e^{1/n}$ to some order ($O(1/n^k)$). Presumably the value of $k$ will be related to the existence of $1/(2n^2)$ in your summand. Then simplify if possible, and see what happens from there. Haven't tried it so can't be 100% sure, but that's what I'd do first. Mar24 comment Complex integration confusion Try $z=e^{i\theta}$, where $\theta\in[0,2\pi]$. Mar21 comment Why Does $\sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k}$ sum to $(1-(1-p)^{n+1})$? The binomial theorem states that: $$(a+b)^n=\sum_{k=0}^n{n\choose k}a^kb^{n-k}.$$ Mar19 comment Having trouble understanding generalized complex numbers @ Vim yes, that's exactly the problem I was thinking might arise. Mar19 comment Having trouble understanding generalized complex numbers @ Vim yes, as I said I'm not too sure about that part as I haven't looked into it a great deal. I think if the discriminant is negative then you still have a linear equation so you could still solve for $i$. My worry was that the $i$ resulting from the $\sqrt{\cdot}$ was a "different" $i$ from that being defined. But as I say I haven't looked into this in any depth whatsoever! Mar19 comment Having trouble understanding generalized complex numbers Just a thought (and might not be applicable here), regarding (1) could you not use the quadratic formula to obtain a "non-recursive" definition, e.g. $$i=\frac{q\pm\sqrt{q^2-4p}}{2}.$$ You would probably need $q^2-4p\geq 0$, although I'm not too sure about that... Mar17 comment Any way to simplify integral of Confluent Hypergeometric Function of the First Kind? @NathanMcKenzie out of interest, from what problem does this integral arise? Mar17 comment Any way to simplify integral of Confluent Hypergeometric Function of the First Kind? You may have better luck using the change of variable $y=\frac{\log n}{t}+1$. This gives $dt=\frac{\log(n)dy}{y^2}$ and $t=-\log n\implies y=0$, $t=0\implies y=\infty$. Then the integral becomes: $$-z\log(n)\int_0^\infty \frac{e^{\frac{\log (n)(1-s)}{y-1}}}{(y-1)^2} {}_1F_1\left(1-z,2,\frac{\log n}{y-1}\right)dy.$$ There are plenty of tables of integrals with limits over $\mathbb{R}^+$. See e.g. Gradshteyn and Ryzhik, p.814. Mar11 comment Is there a name for an object with both position and velocity? @Sebastian, so for example a point in phase space might be $p=(x,y,u,v)$, where $(x,y)$ is position and $(u,v)$ is velocity. Thank you. Feb26 comment Does the integral $\int_{1}^{\infty} \sin(x\log x) \,\mathrm{d}x$ converge? I understand you're using the Lambert W function which enables you to obtain $x=g(y)$, but could you please explain in a little more detail how you got the substituted integral? Feb19 comment What is known about the complex solutions to $\zeta(s)=-1$? I suppose this is related to "level curves" of a function...