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Questions tagged [gaussian-integral]

For questions regarding the theory and evaluation of the Gaussian integral, also known as the Euler–Poisson integral is the integral of the Gaussian function $~e^{−x^2}~$ over the entire real line. . It is named after the German mathematician Carl Friedrich Gauss.

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1 answer
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Consider the following integral: $$I=\int_0^\infty dx\,e^{-x^2\frac{1+j}{\sqrt{2}}}.$$ where j is the imaginary unit. We get: $$I^2=\int_0^\infty \int_0^\infty dx dy e^{-(x^2+y^2)\frac{1+j}{\sqrt{2}}}....
Aria's user avatar
  • 442
0 votes
2 answers
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Let $I$ be the characteristic function $$I(x) = \begin{cases} 1 \quad |x| \leq c \\ 0 \quad |x| > c\end{cases}$$ and let $\mu$ be a standard Gaussian measure on $\mathbb{R}^n$. How can I prove that ...
Mathematics's user avatar
2 votes
1 answer
106 views

I am working through Chapter 6 of the book Statistical Mechanics of Machine Learning by Engel and Van den Broeck. I am stuck on the following integral, going from line 6.13 to line 6.14 of the book. I ...
c_lienk's user avatar
  • 100
2 votes
0 answers
141 views

For $n \in \mathbb{N}$, let $f:[0,1]^n \to \mathbb{R}_+$ continuous. For some $t \in \mathbb{R}$ we want to compute: $$ I(t) = \int_{[0,1]^n} dx \cdot e^{-t^2 \cdot f(x)} $$ One can have the following ...
C Marius's user avatar
  • 1,505
1 vote
1 answer
82 views

I have a double integral, which I want to solve: $\mathcal{I}_{au}=\int_0^{t'}e^{(a-c)t''}\int_0^{t''}e^{(-b+c)t'''}\text{exp}\big(-\frac{\lambda^2}{2}\left( S_2(t''')^2+S_1(t')^2+S_1(t'')^2+2S_{12}t'...
Nick Andersson1411's user avatar
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0 answers
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This is a generalization of another question from here.. Let $n \ge 2$ be an integer. Let $\vec{a}:=(a_i)_{i=1}^n \in {\mathbb R}_+^n$ and ${\bar a}:=(a_{i,j})_{1\le i < j \le n} \in {\mathbb R}_+^...
Przemo's user avatar
  • 12.1k
1 vote
1 answer
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Let $H \in (0,1)$ and let $\delta \ge 0 $ and let $d \in {\mathbb N}_+$. We are interested in computing a probability that, in a fractional Brownian Noise with Hurst exponent $H$ there are $n$ ...
Przemo's user avatar
  • 12.1k
3 votes
1 answer
111 views

Consider the following integral over $\mathbb{R}^{q+1}$ $$I_q(s) := \int_{\substack{y_1, \ldots, y_q > s \\ y_{q+1} <s}} \exp\left(-Y\Sigma Y^T\right) d\lambda(Y),$$ where $Y = (y_1, \ldots, y_{...
hseldon39's user avatar
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0 answers
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In my atomic physics research, I encountered the following operator: $$G(v_1,v_2)=(\mathbf{1}+v_1L_1+v_2L_2)^{-1}=\dfrac{1}{\mathbf{1}+v_1L_1+v_2L_2}$$ where $v_{1,2}$ are Gaussian random variables ...
Omar Nagib's user avatar
  • 1,326
0 votes
2 answers
77 views

I need help, "solving" the following integral. \begin{align} \int_{0}^{\infty}\exp\left(-a(x-b)^2\right)\; dx \end{align} Does this integral converge? In my opinion it does, because the form ...
I8SumPi's user avatar
  • 99
-4 votes
1 answer
84 views

A standard result for the Gaussian integral is that $\int_{\infty}^{\infty} dx \: e^{- x^2} = \sqrt{\pi}$. If instead, one has $ \int_{\infty}^{\infty} dx \: e^{-\frac{(x-c)^2}{\sigma^2}} $ for ...
Tom's user avatar
  • 3,135
3 votes
2 answers
363 views

I recently stumbled upon a problem I solved in the past and realized my steps were wrong but I got the right answer. In the past after finding the gaussian integral, I tried applying the same steps in ...
MyDogLovesVR's user avatar
3 votes
1 answer
178 views

I am trying to evaluate the following integral: $$I = \int_{-\infty}^{\infty} 6x e^{-3x^2} e^{-i\alpha x} \,\mathrm dx.$$ To simplify this, I first complete the square in the exponent. The exponent in ...
Pizza's user avatar
  • 377
0 votes
1 answer
65 views

Let $f:\mathbb{R}^n \to \mathbb{R}$ be differentiable and Lipschitz continuous (i.e., $\|\nabla f(x)\| \le L$ for any $x \in \mathbb{R}^n$). Furthermore, define the Gaussian-smoothed function $f_\mu$ ...
yu h's user avatar
  • 13
2 votes
3 answers
187 views

I am trying to find a version of the following integral in terms of special functions: $$ \int _{\alpha }^{\beta }\frac{e^{-r^2 \sigma ^2} \text{erf}(\gamma \sigma )}{\sigma }d\sigma \quad (1) $$ for ...
Idividedbyzero's user avatar

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