Reputation
1,245
Top tag
Next privilege 2,000 Rep.
Edit questions and answers
Badges
1 13 29
Impact
~70k people reached

2d
accepted Solving the IVP given by $\dot x=\frac{t-x}{t+x}$ and $x(0)=1$
2d
comment Solving the IVP given by $\dot x=\frac{t-x}{t+x}$ and $x(0)=1$
I don't seem to understand your notation which is different than everything I have ever seen. Could you explain more thoroughly?
2d
comment Solving the IVP given by $\dot x=\frac{t-x}{t+x}$ and $x(0)=1$
Applying your substitution yields $\dot v = \frac{(1-v)}{(1+v)}$ but I can't seem to continue from there.
2d
asked Solving the IVP given by $\dot x=\frac{t-x}{t+x}$ and $x(0)=1$
May
2
comment Gaussian integral $\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)\,\mathrm d x$ along $[-R,R]+\mathrm i[0,Y]$
Your solution is very thoroughly and not messing around too much with curves as I did - I like your solution. Thank you very much for your input!
May
2
accepted Gaussian integral $\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)\,\mathrm d x$ along $[-R,R]+\mathrm i[0,Y]$
May
2
accepted Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?
May
2
comment Gaussian integral $\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)\,\mathrm d x$ along $[-R,R]+\mathrm i[0,Y]$
And finally, how can you exactly justify the last step in $(2)$ where you replaced $z$ with $x+\mathrm iY$ as well as the boundaries of the integral? I assume one can argue, it is the result of a substitution $z=x+\mathrm iY$ hence reversing it yields the desired result.
May
2
comment Gaussian integral $\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)\,\mathrm d x$ along $[-R,R]+\mathrm i[0,Y]$
So If I understand your approach correctly, you don't start at the integral over $\exp(-(x+\mathrm iY)^2)$ but with the Gaussian integral and then you develop which curves are necessary, isn't it? If I substitute my second curve (from $R$ to $R+\mathrm iY$) in the way to imply it I would get $$\int_{R+\mathrm iY}^{R+2\mathrm iY}\exp(-u^2)\,\mathrm du$$ which is bad since the upper boundary isn't nice.
May
1
revised Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?
added own solution for follow-up question
May
1
comment Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?
Please refer to my edited question with a second question.
May
1
comment Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?
Please refer to my edited question with a second question.
May
1
revised Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?
added a follow-up question
May
1
comment Gaussian integral $\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)\,\mathrm d x$ along $[-R,R]+\mathrm i[0,Y]$
@πr8 My last comment was actually false. Letting $u=R(1-2t)+2\mathrm iY$ for the third curve yields $$\int_{R+2\mathrm iY}^{-R+2\mathrm iY}\exp(-u^2)\,\mathrm du$$ while the substitution $u=R(1-2t)$ yields your suggested boundaries with a wrong integrand namely $$\int_{R}^{-R}\exp(-(u+2\mathrm iY)^2)\,\mathrm du.$$ Is it reasonable to take the first substitution I mentioned here show the equality and argue that it doesn't matter that the boundaries are shifted by $2\mathrm iY$ since this is a constant which doesn't matter for $R\to\infty$?
May
1
comment Gaussian integral $\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)\,\mathrm d x$ along $[-R,R]+\mathrm i[0,Y]$
@πr8 Since the integrand has an antiderivative it follows from the theorem that $$\int_{-R}^R \exp(-(u+\mathrm iY)^2)\,\mathrm du - \int_{-R+\mathrm iY}^{R+\mathrm iY} \exp(-(u+\mathrm iY)^2)\,\mathrm du = 0$$ and thus both integrals are equal. My issue right now is to find the connection to $\int_{-\infty}^\infty\exp(-x^2)\,\mathrm dx$. Surely it is some simple integration rule, isn't it?
May
1
comment Gaussian integral $\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)\,\mathrm d x$ along $[-R,R]+\mathrm i[0,Y]$
@πr8 Alright the behaviour of the vertical lines is indeed obvious thus (omitting said vertical lines since their integrals vanish for $R\to\infty$) $$\oint_{\partial Q}\exp(-(x+\mathrm iY)^2)\,\mathrm dx = \int_{-R}^R\exp(-(u+\mathrm iY)^2)\,\mathrm du + \int_{R+\mathrm iY}^{-R+\mathrm iY}\exp(-(u+\mathrm iY)^2)\,\mathrm du.$$ However I am still unsure about deducing something from this in regards to the Gaussian integral.
May
1
comment Gaussian integral $\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)\,\mathrm d x$ along $[-R,R]+\mathrm i[0,Y]$
@πr8 I edited the question to add my calculations thus far - please have a look at it as I haven't found too many similiarities besides the boundaries of the integrals and start/end point of the curves.
May
1
revised Gaussian integral $\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)\,\mathrm d x$ along $[-R,R]+\mathrm i[0,Y]$
added some detailed results of own computations
May
1
asked Gaussian integral $\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)\,\mathrm d x$ along $[-R,R]+\mathrm i[0,Y]$
May
1
comment Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?
So far we haven't been talking about homotopy so I find it difficult to fully understand your proof. I have a feeling you are trying to pick arbitrary curves which start and end point are fixed and showing that the integrals along those curves is always the same. Can you confirm this? I am struggling to understand how you deduce this exactly.