# Tag Info

21

Here is my trial, which is partially successful but still not fully answering to your question. Using some coordinate change, I derived that $$I_{n} := \int_{0}^{\infty} \mathrm{erfc}^{n}(x) \, dx = \frac{1}{\sqrt{n}} \left( \frac{2}{\sqrt{\pi}} \right)^{n-1} \int_{T^{n-1}} \int_{0}^{\infty} s^{n-1}e^{-(|x|^{2}-1)s^{2}} \mathrm{erfc}(s) \, ds d\sigma_{x}, ... 19 First I'm going to integrate by parts twice.$$\int_{0}^{\infty} \text{erfc}^{3}(x) \ dx = x \text{erfc}^{3}(x) \Big|^{\infty}_{0} + \frac{6} {\sqrt{\pi}} \int_{0}^{\infty}x \ \text{erfc}^{2}(x) e^{-x^{2}} \ dx = - \frac{3 \ \text{erfc}(x) e^{-x^{2}}}{\sqrt{\pi}} \Big|_{0}^{\infty} - \frac{12}{\pi} \int_{0}^{\infty}\text{erfc}(x) e^{-2x^{2}} \ dx$$... 14 Not sure how to transform conjecture (2) to (5). However, (5) is true. Let \;\displaystyle t = \frac{1}{1+y^3}\;, we can rewrite the integral \;\displaystyle B\left(\frac19;\frac13,\frac16\right)\; as$$ \int_0^{1/9} t^{-5/6} (1-t)^{-2/3} dt = \int_\infty^2 (1+y^3)^{5/6}\left(\frac{1+y^3}{y^3}\right)^{2/3}\frac{-3y^2 dy}{(1+y^3)^2} = 3 ...

14

$\int^{\pi}_0\cos^{2k+1}\phi d\phi=0$, and $$\int^{\pi}_0\cos^{2k}\phi d\phi=\sqrt{\pi}\Gamma(k+1/2)/\Gamma(k+1)=2^{-2k}\pi\binom{2k}{k}.$$ Therefore \begin{align*} I(n) &=\int^{\pi}_0\sum_{m=0}^{\infty}2^{n-m}\binom{n}{m}\cos^m\phi~d\phi\\ &=2^n\sum_{m=0}^{\infty}2^{-m}\binom{n}{m}\int^{\pi}_0\cos^m\phi~d\phi\\ ... 9 Using Euler-type integral representation for the Gauss's function: {}_2F_1(a,b; c; z) = \frac{\Gamma(c)}{\Gamma(b) \Gamma(c-b)} \int_0^1 u^{b-1} (1-u)^{c-b-1} (1-z u)^{-a} \mathrm{d}u $$for c = \tfrac{4}{3} and b=\tfrac{1}{3}, differentiating with respect to a at a=\tfrac{5}{6}:$$ \left.\frac{\mathrm{d}}{\mathrm{d}a} ...

9

Well you can do $$\int_0^{\pi}{\sin(\sin x)} = \pi H_0(1)\,,$$ where $H$ is a non-elementary function known as the Struve function. (For info on Struve functions, see http://mathworld.wolfram.com/StruveFunction.html.) You might be wondering...what good does it do to express the integral in terms of a function that almost nobody has ever heard of? Well, ...

8

We have $$\int_{-\infty}^{\infty} \operatorname{erfc}^{2}(|x|) e^{-i\xi x} \, dx = \frac{4}{\xi}e^{-\xi^{2}/4} \left\{ \operatorname{erfi}\left( \frac{\xi}{2} \right) - \operatorname{erfi}\left( \frac{\xi}{2\sqrt{2}} \right) \right\}, \tag{1}$$ where $\operatorname{erfi}$ is the imaginary error function defined by $$\operatorname{erfi}(x) = ... 7 For s=0,1,2,\ldots we compute$$ \mathcal{J}(s) = 1, 0, 1, 5, 36, 329, 3655, 47844, 721315, 12310199, 234615096, \ldots $$[This was obtained with the gp code F(p) = sum(k=0, poldegree(p), k!*polcoeff(p,k)) N = 10 vector(N+1,s,F(pollegendre(s-1,x-1))) and can also be obtained from the expansion of P_s(x-1) in powers of x, giving (-1)^s ... 6 When 0<x<1, we have the identity K(x^{-1})=x(K(x)-iK(\sqrt{1-x^2})). Therefore$$\int^{\infty}_{1}K^2(x)\frac{dx}{x}=\int^{1}_{0}K^2(x^{-1})\frac{dx}{x}\\ =\int^{1}_{0}\left(x(K(x)-iK(\sqrt{1-x^2}))\right)^2\frac{dx}{x}\\ =\int^{1}_{0}x\left(K(x)-iK(\sqrt{1-x^2})\right)^2dx\\ =\int^1_0(xK^2(x)-xK^2(\sqrt{1-x^2})-2ixK(x)K(\sqrt{1-x^2}))dx.$$We ... 5 Following my answer in this question, we have$$ \begin{align*}&\; \int^1_0x^nK\left(\sqrt{\vphantom1x}\right)K\left(\sqrt{1-x}\right)dx\\ &=\int^1_0\left(\frac\pi2\sum^{\infty}_{m=0}\frac{(2m)!^2}{2^{4m}(m!)^4}x^{m+n}\right)K\left(\sqrt{1-x}\right)dx\\ ...

5

Since the substitution $s = (1+k)x$ transforms $$\int_a^b x^\alpha e^{-(1+k)x}\,dx$$ into $$\frac{1}{(1+k)^{\alpha+1}} \int_{(1+k)a}^{(1+k)b} s^\alpha e^{-s}\,ds,$$ we can't pull the integral out of the sum as a constant, so the argument from a compact subinterval of $(0,\infty)$ does at least not work too easily. The simplest way would of course be to ...

5

I reduced $I_5$ to a simply looking integral of elementary functions. The statement is: \begin{eqnarray} I_5&=&\int_0^{\infty}\operatorname{erfc}^5x\,dx =\\ &=&\frac{5}{\pi^{1/2}}-\frac{240\sqrt{2}\arctan{\sqrt{2}}}{\pi^{5/2}}\left(\arctan2- \frac{\pi}{4}\right)+\frac{480}{\pi^{5/2}}\int_{\operatorname{arcsinh 1}}^{\infty}\frac{\arctan(\cosh ...

4


4

By "anywhere in literature" did you include L. Lewin, Polylogarithms and Assoicated Functions ? In the (fortcoming) solution for Monthly problem 11654, you will see that very equation. For the proof, our solver (Richard Stong) used these identities: \mathrm{Li}_3\left(\frac{1-z}{1+z}\right) - \mathrm{Li}_3\left(\frac{z-1}{z+1}\right) = ... 4 Combining trilogarithm identities 1 and 2, one obtains the formula \begin{align} \operatorname{Li}_3\left(\frac{1-z}{1+z}\right)-\operatorname{Li}_3\left(-\frac{1-z}{1+z}\right)= 2\operatorname{Li}_3\left(1-z\right)+2\operatorname{Li}_3\left(\frac{1}{1+z}\right)- ... 3 The proof below uses only the chain rule (to change integration variables). It's what I, as a physics student, would write. But first, I believe that you have an error in your original posting. I believe that the 1D identity should be: \begin{align} \int \Big( \sum_{\substack{\textrm{roots a_i}\\ \textrm{ of g}}} \frac{1}{g'(a)} \Big)^{-1} ... 3 ax^{2} + bx = a\left[\left(x-\frac{b}{2a}\right)^{2} - \frac{b^{2}}{4a^{2}}\right] $$make the substitution$$ z = \sqrt{2a}\left(x -\frac{b}{2a}\right) $$you will end up with an integral like this$$ \frac{\mathrm{e}^{\frac{b^{2}}{4a}}}{2a}\int \left(z-\frac{b}{\sqrt{2a}}\right)\mathrm{e}^{-\frac{z^{2}}{2}}dz $$or if we split it up$$ ...

3

I'm assuming you mean $$\zeta(z) = \frac{\Gamma(1-z)}{2 \pi i} \int_{C} \frac{t^{z-1} }{e^{-t}-1} \ dt = \frac{\Gamma(1-z)}{2 \pi i} \int_{C} \frac{t^{z-1}e^{t} }{1-e^{t}} \ dt$$ where $C$ is a contour on the complex plane that starts at $- \infty$ below the branch cut on the negative real axis, goes around the origin, and then goes to back to $-\infty$ ...

2

The integration problems one gets in Calculus 2 have integrands specially tailored to admit sufficiently simple closed-form antiderivatives. In the platonic world of mathematics, integrands are more likely to have horrendous-looking antiderivatives than clean-looking, and much more likely still to not have any elementary closed-form. We have to be picky ...

2

The integral condition says very little about $f$; it selects an affine subspace of codimension $1$, which is not any more manageable than the space you began with. For any $f\in L^1(\mathbb R)$, the function $$f-c\chi_{[0,1]},\quad \text{where } c = \int_{-\infty}^\infty f -1$$ satisfies your condition. So, if you have a convenient representation here, ...

2

The constant functions $C(x)=1$ and $S(x)=0$ satisfy these conditions. Now that you added the condition of periodicity exactly $2\pi$, the only continuous spurious solution is $C(x)=\cos(x)$ and $S(x)=-\sin(x)$. However there are still plenty of other discontinuous solutions. Here is why. First note that if one would have $C(x)=S(x)=0$ for any particular ...

2

$$\int_{0}^{\infty} F(x) F(x \sqrt{2}) \frac{e^{-x^{2}}}{x^{2}} dx = \int_{0}^{\infty} \int_{0}^{\sqrt{2}} \int_{0}^{1} x e^{-x^{2}} e^{x^{2} y^{2}} x e^{-2x^{2}} e^{x^{2}z^{2}} \frac{e^{-x^{2}}}{x^{2}} \ dy \ dz \ dx$$ $$= \int_{0}^{\sqrt{2}} \int_{0}^{1} \int_{0}^{\infty} e^{-(4-y^{2}-z^{2})x^{2}} \ dx \ dy \ dz = \frac{\sqrt{\pi}}{2} ... 2 Following the analysis here:$$\begin{align} F(x) &= \int_0^{\infty} \frac{dy}{y} e^{-y-\frac{x}{y}}\end{align}$$Sub u=y+\frac{x}{y}, then$$y = \frac12 \left (u \pm \sqrt{u^2-4 x}\right )dy = \frac12 \left ( 1 \pm \frac{u}{\sqrt{u^2-4 x}} \right ) du$$Then$$\begin{align}F(x) &= \frac1{4 x} \int_{\infty}^{2 \sqrt{x}} du \left ( 1 - ...

2

It follows from the duplication formula for the gamma function. $$B(z,z) = \frac{\Gamma(z) \Gamma(z)}{\Gamma(2z)} = \Gamma(z) \Gamma(z) \frac{\sqrt{\pi}}{2^{2z-1} \Gamma(z) \Gamma (z+1/2)}$$ $$= 2^{1-2z} \frac{\Gamma (z) \Gamma(1/2)}{\Gamma(z+1/2)} = 2^{1-2z}B \left(z, \frac{1}{2} \right)$$ http://mathworld.wolfram.com/LegendreDuplicationFormula.html

2

One possibility is the following: Replace $$\frac{1}{k^2+a^2}=\int_0^{\infty}e^{-s(k^2+a^2)}ds.$$ The integral over $n$ components of $k$ now factorizes into the product of $n$ Gaussian integrals ...

2

Incomplete solution: In fact, we have $${}_2F_1(-\frac14,\frac54;1;\frac{x}{2})=\frac{8\sqrt{2}}{\pi\sqrt{2+\sqrt{2x}}}((2+\sqrt{2x})E(\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}})-K(\frac{2\sqrt{x}}{\sqrt{2}+\sqrt{x}})).$$ Here $E(m)=\int^1_0\sqrt{\frac{1-mt^2}{1-t^2}}~dt$ and $K(m)=\int^1_0\frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}$ are the complete elliptic integrals ...

2

I think the series definition is best. The series definition of the Bessel function is $$J_{\nu}(z) = \sum_{n=0}^{\infty}\frac{(-1)^n}{\Gamma(n+\nu+1)\,n!}\left(\frac{x}{2}\right)^{2n+\nu}.$$ We will expand $e^{iz\cos(\theta)}$ as a power series as follows: $$e^{iz\cos(\theta)} = \sum_{n=0}^{\infty}\frac{1}{n!}(iz\cos(\theta))^n.$$ So integrating term by ...

2

Here is an outline of a proof: First use the integral definitions to show that $$k \frac{d E(k)}{dk} = E(k)-K(k)$$ and $$k(k^{*})^{2} \frac{d K(k)}{d k} = E(k) - (k^{*})^{2} K(k) .$$ Then use those identities to show that $$\frac{d}{dk} \left( K(k) E(k^*)+ E(k) K(k^*) - K(k) K(k^*) \right) =0 .$$ Then integrate back and take the limit as $k$ goes to ...

1

For $\psi ^{(0)}(x)$, Taylor expansion around $x=0$ is given by $$-\frac{1}{x}-\gamma +\frac{\pi ^2 x}{6}+\frac{x^2 \psi ^{(2)}(1)}{2}+\frac{\pi ^4 x^3}{90}+\frac{x^4 \psi ^{(4)}(1)}{24}+\frac{\pi ^6 x^5}{945}+O\left(x^6\right)$$ Apply it to the expression and the final result is $$-\frac{\pi ^2 x}{12}$$

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