Question

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Answer 1

So what have you tried?

Answer 2

Gaussians are always fun xD

Answer 3

\[\int\limits_0^{\infty} e^{-st}e^{(t-1)^2}dt = \int\limits e^{-st} e^{t^2 - 2t + 1}dt = \int\limits e^{t^2 - (s+2)t+1}dt\]

Answer 4

How did you approach that integral?

Answer 5

I like your questions just btw xP

Answer 6

It does diverge, if you consider complex numbers you can get this answer https://www.wolframalpha.com/input/?i=laplace+transform+of+e^%28%28t-1%29^2%29
Which doesn't have much meaning to me honestly. I did it on paper and it does actually diverge.

Answer 7

The general idea of gaussians is you have:
\[\int\limits e^{\alpha x^2 + \beta x + \gamma}dx\]
You want to complete the square and get it in a form:
\[\int\limits e^{-\xi u^2}du\] and (depending on the bounds) you know what this is. For example:
\[\int\limits_{-\infty}^{\infty} e^{- \xi x^2}dx = \sqrt{\frac{\pi}{\xi}}\]

Answer 8

I believe because one of the conditions is that the integral must exist.

Answer 9

This is what I did using that technique. But the integral does diverge.

Answer 10

Something even more interesting is if you have:
\[\underbrace{\int\limits_{-\infty}^{\infty}...\int\limits_{-\infty}^{\infty}}_{\textrm{'n integrations'}}\exp \left[ (x_1,x_2,...,x_n) A (x_1,x_2,x_3,...,x_n)^T\right]dx_1 \wedge dx_2 \wedge dx_3...\wedge dx_n\]
That is given by:
\[\left(\frac{\pi}{\det(A)}\right)^\frac{n}{2}\]

Answer 11

Sorry but I have to go D:

I'll be back at some point though.