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39

Another proof, from G.M. Fichtengoltz, Calculus Course, page 612. $$K=\int_{0}^{\infty} e^{-x^2} dx$$ It easy to see (and prove) that, $\max\{(1+t)e^{-t}\}=1$ at $t=0$, hence for all $t\in\mathbb{R}$: $$(1+t)e^{-t}<1$$ Substitution of $t=\pm x^2$, leads us to: $$(1-x^2)e^{x^2}<1 \ \ \ \ \text{and} \ \ \ \ \ (1+x^2)e^{-x^2}<1$$ So, $${1-x^... 31 The presentation here is typical of those used to model and motivate the infinite dimensional Gaussian integrals encountered in quantum field theory. I will use subscripts instead of superscripts to indicate components. I. Wick's theorem First consider the integral$$Z_0 = \int d^n x \exp\left(-\frac{1}{2} x^\mathrm{T} A x\right),$$where d^n x = \prod_i ... 29 By using Beta and Gamma functions properties we may simply obtain that:$$\operatorname B\left(\tfrac 12,\tfrac12\right)=\frac{\left[\Gamma(\tfrac{1}{2})\right]^{2}}{\Gamma{(1)}}=\left[\Gamma(\tfrac{1}{2})\right]^{2}\operatorname B\left(\tfrac{1}{2},\tfrac{1}{2}\right)=\frac{\pi}{\sin{\frac{\pi}{2}}}=\pi$$In other words we have that:$$\Gamma(\tfrac{1}{...

23

First, we notice that $$n!\ =\ \int_0^\infty e^{-\sqrt[n]x}\ dx\quad\iff\quad\tfrac1n!\ =\ \int_0^\infty e^{-x^n}dx\quad\rightarrow\quad\tfrac12!\ =\ \int_0^\infty e^{-x^2}dx$$ Then we further notice that $$\int_0^1\Big(1-\sqrt[n]x\Big)^m\,dx\ =\ \int_0^1\Big(1-\sqrt[m]x\Big)^n\,dx\ =\ \frac1{C_{m+n}^n}\ =\ \frac1{C_{m+n}^m}\ =\ \frac{m!\,n!}{(m+n)!}$$ From ...

18

The following argument, similar to Bryan Yock's, is a Feynman parameter trick I invented in Integrating $\int^{\infty}_0 e^{-x^2}\,dx$ using Feynman's parametrization trick Let $$I(b) = \int_0^\infty \frac {e^{-x^2}}{1+(x/b)^2} \mathrm d x = \int_0^\infty \frac{e^{-b^2y^2}}{1+y^2} b\,\mathrm dy$$ so that $I(0)=0$, $I'(0)= \pi/2$ and $I(\infty)$ is the ...

13

Hint. Make the change of variable $t=x^2$ to obtain $$\int_0^{\infty}\frac{e^{-(t+\frac{1}{t})}}{\sqrt t}dt=2\int_0^{\infty}e^{ -x^2-1/x^2}dx=\int_{-\infty}^{\infty}e^{ -x^2-1/x^2}dx$$ You may then recall that, for any integrable function $f$, we have \int_{-\infty}^{+\infty}f\left(x-\frac{s}{x}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \... 12 Consider \begin{align} x^{2} + \frac{1}{x^{2}} = \left( x - \frac{1}{x} \right)^{2} +2 \end{align} for which \begin{align} I = \int_{0}^{\infty} e^{-\left(x^{2} + \frac{1}{x^{2}}\right)} \, dx = e^{-2} \, \int_{0}^{\infty} e^{-\left(x - \frac{1}{x}\right)^{2}} \, dx. \end{align} Now make the substitution t = x^{-1} to obtain \begin{align} e^{2} I = \int_{... 12 Change variables. Let z=x^2. We find \int_{0}^{\infty} e^{-x^2} dx = \frac{1}{2} \Gamma(\frac{1}{2}) = \frac{\sqrt{\pi}}{2}. Addendum: Setting z=1/2 in Euler's reflection formula, \Gamma(1-z)\Gamma(z) = \pi/\sin \pi z, we find \Gamma(1/2) = \sqrt{\pi}. 12 If you consider every possible outcome of some event you should expect the probability of it happening to be 1, not \sqrt{2\pi} so the constant scales the distribution to conform with the normal convention of ascribing a probability between zero and one. 9 Here you have a solution using Wallis' formula for \pi and the squeeze theorem which I learned. It's from a post I made in a forum. http://www.mymathforum.com/viewtopic.php?f=15&t=27064 9 Differentiation under the integral sign gives a pretty fast way. Let: I(\alpha) = \int_{0}^{+\infty}x^{\alpha}e^{-x^2}\,dx = \frac{1}{2}\int_{0}^{+\infty}x^{\frac{\alpha-1}{2}}e^{-x}\,dx = \frac{1}{2}\,\Gamma\left(\frac{\alpha+1}{2}\right).\tag{1}$$Our integral is just I'(0): since \Gamma' = \psi\cdot\Gamma,$$ \int_{0}^{+\infty}e^{-x^2}\log(x)\,dx =...

8

Consider the mapping $\eta\! : \mathbb{R}^2\to \mathbb{R}$ given by $$\eta((x,y)) = \sqrt{x^2+y^2},\quad (x,y)\in\mathbb{R}^2.$$ (1) Show that the image-measure $\lambda_2\circ\eta^{-1}$ is the measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ with density $$f(z)=2\pi z 1_{(0,\infty)}(z),\quad z\in\mathbb{R},$$ by using Dynkin's Lemma. (2) Show that $$\... 8 Integration by parts does help! Let's assume a>0, n\in\mathbb{N}. Call your integral I_{2n}. Then,$$ \begin{align} I_{2n}&=\int_{-\infty}^\infty e^{-\frac{1}{2}a x^2} x\cdot x^{2n-1} dx=\int_{-\infty}^\infty -\frac{1}{a} \frac{d}{dx}\left(e^{-\frac{1}{2} ax^2}\right) x^{2n-1} dx \\ &= \frac{2n-1}{a} I_{2n-2} \end{align} $$Knowing that ... 8 So we want to compute$$\oint_C dz \frac{e^{i \pi z^2}}{\sin{\pi z}} $$where C = C_1+C_2+C_3+C_4. Along C_1, z=-1/2 + e^{i \pi/4} t, t \in [R,-R].$$\begin{align}\int_{C_1} dz \frac{e^{i \pi z^2}}{\sin{\pi z}} &= -e^{i \pi/4} \int_{-R}^R dt \, \frac{e^{i \pi (-1/2+e^{i \pi/4} t)^2}}{\sin{\pi (-1/2+e^{i \pi/4} t)}}\\ &=i \int_{-R}^R dt \,\...

8

This is Gauss' Linking Number Formula, for two space curves $\vec{A}, \vec{B}: S^1 \to \mathbb{R}^3$ $$\textrm{link}(A,B) = \oint_A \oint_B \frac{\vec{A}-\vec{B}}{|\vec{A}-\vec{B}|^3} \cdot (d\vec{A} \times d\vec{B})$$ In our case, $\vec{A}(t) = (\cos t, \sin t, 0)$ and $\vec{B}(t) = ( 1+ \cos 2t, \frac{1}{2}\sin t, \sin 2t)$ . How to picture these two ...

8

In the last equality of the second equation, you tried to utilize Fubini's theorem. But the issue is that the integral is only conditionally convergent. So in principle, you cannot apply Fubini's theorem to convert your iterated improper integral into the double integral. Assuming that you have already established the existence of $I$, here are two ...

8

Here is a way forward that provides a solution in terms of a finite double summation. Let $f(a)$ be defined by $$f(a)=\int_{-\infty}^\infty \frac{e^{-ax^2}}{(1+x^2)^2}\,dx=\sqrt{\pi a}-\frac{\pi}{2}(2a-1)e^a \text{erfc}(\sqrt {a}) \tag 1$$ Then, the $n$'th derivative of $f(a)$ can be written f^{(n)}(a)=\int_{-\infty}^\infty \frac{(-1)^nx^{2n}e^{-ax^... 7 Since the integrand is even then \begin{align} \int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx=2\int_{0}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx \end{align} Using substitution t=x^2 we get \begin{align} \int_{-\infty}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx&=2\int_{0}^\infty e^{- \frac{1}{2} ax^2 } x^{2n}\, dx\\ &=\int_{0}^\infty t^{... 7 This way is easier than IBP: \begin{align} & \hspace{6mm}\int_{-\infty}^\infty e^{- \lambda x^2 } x^{2n} dx \\ &= \int_{-\infty}^\infty (-1)^n \frac{d^n}{d \lambda^n} e^{-\lambda x^2} dx \\ &= (-1)^n \frac{d^n}{d \lambda^n} \int_{-\infty}^\infty e^{-\lambda x^2} dx \end{align} 7 A corrected form of the question asks to show that \int_{\mathbb R^n} e^{-x^tAx}\;dx\;=\; \pi^{n/2}/\sqrt{\det A} for symmetric n-by-n A with positive-definite real part. First, for A real (positive-definite), there is a (unique) positive-definite square root S of A, and the change of variables x=S^{-1}y gives the result, as the questioner ... 7 The Euler-MacLaurin summation formula can be used to approximate n! conveniently. The following is taken from Concrete Mathematics by D.E. Knuth ad R.L. Graham. We find in section 9.5 among other representations this version of the Euler-MacLaurin summation formula for integer values a\leq b \begin{align*} \sum_{a\leq k < b}f(k)&=\int_a^bf(x)... 7 One typical way to write integration by parts is to identify your original integral as \int u \, dv and then we can use the identity\int u \, dv = uv - \int v \, du$$Where we form du from u by differentiating u with respect to the independent variable, and we form v from dv by antidifferentiating. In your case, if I am taking you rightly you ... 7 Let g(x) be a primitive of e^{-x^2}. Then$$\lim_{x\to 0}\frac{g(2x)-g(x)}x=\lim_{x\to 0}\left(2\frac{g(2x)-g(0)}{2x}-\frac{g(x)-g(0)}x\right)=2g'(0)-g'(0)=1.$$7 It's better to simplify the term \mathbb{E}[(x-\mu)^T \Sigma^{-1}(x-\mu)] directly:$$ \begin{align} \mathbb{E}[(x-\mu)^T \Sigma^{-1}(x-\mu)] &= \mathbb{E}[\mathrm{tr}((x-\mu)^T \Sigma^{-1}(x-\mu))]\\ &= \mathbb{E}[\mathrm{tr}(\Sigma^{-1}(x-\mu)(x-\mu)^T)]\\ &= \mathrm{tr}(\mathbb{E}[\Sigma^{-1}(x-\mu)(x-\mu)^T])\\ ...

7

With a substitution the evaluation of $(1)$ boils down to the evaluation of $$3\int_{0}^{+\infty} x e^{-x} \log(1+x^{1/3})\,dx \stackrel{IBP}{=} \int_{0}^{+\infty}\frac{x+1}{x^{2/3}+x}e^{-x}\,dx\tag{A}$$ and with the inverse substitution the RHS of $(A)$ turns into $$3\int_{0}^{+\infty}(x^2-x+1)\,e^{-x^3}\,dx = \Gamma\left(\frac{3}{3}\right)-\Gamma\left(\... 7 Starting with the generating function$$ e^{2xt-t^2}=\sum_{n\geq 0}H_n(x)\frac{t^n}{n!}\tag{1} $$then replacing t with t e^{i\theta} we have$$ \exp\left[2xt e^{i\theta}-t^2 e^{2i\theta}\right] = \sum_{n\geq 0}H_n(x) e^{ni\theta}\frac{t^n}{n!}\tag{2} $$and by Parseval's identity$$ \int_{-\pi}^{\pi}\exp\left[4xt\cos\theta-2t^2\cos(2\theta)\right]\,d\...

7

This approach doesn't work. The set $\{re^{i\theta} : 0\leq r \leq a, \theta \in [0,2\pi]\}$ is not the same as $\{x+iy, -a\leq x \leq a, -a\leq y \leq a\}$. The first one a disk and the second one a rectangle.

6

Consider the integral: $$\int_0^\infty t^{-1/2}{e^{-t}} dt$$ We perform a change of variables $u=t^{1/2}$ and $du= \frac{1}{2}t^{-1/2} dt$. The integral then becomes: $$\int_0^\infty t^{-1/2}{e^{-t}} dt=\int_0^\infty 2{e^{-u^2}} du.$$ Now let us consider the well-known integral: $$\frac{\pi}{2}=\int_0^\infty \frac{1}{1+x^2} dx$$ We can expand the ...

6

Thinking $L^1_{loc}\cap\mathcal{S}'$ as a subset of $\mathcal{D}'$, it is not true that every element of $L^1_{loc}\cap\mathcal{S}'$ is a polynomially growing function. For example, in $\mathbb{R}$, define $$f:\mathbb{R}\to\mathbb{R}, t\mapsto \cos(e^t) e^t.$$ Then $f\in L^1_{loc}(\mathbb{R})$, so it represents by integral pairing an element of \$\mathcal{...

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