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## Hot answers tagged definite-integrals

15

Sub $x=e^{-u}$, $dx = -e^{-u} du$. Then the integral is $$\int_0^1 dx \frac{x^3-x^2}{\log{x}} = \int_0^{\infty} du \, \frac{e^{-3 u} - e^{-4 u}}{u} = \int_0^{\infty} du \, \int_3^4 dt \, e^{-u t} \\ = \int_3^4 dt \,\int_0^{\infty} du \, e^{-u t} = \int_3^4 \frac{dt}{t} = \log{\frac{4}{3}}$$ The change in the order of integration is justified by Fubini's ...

15

This integral can be evaluated in a closed form for arbitrary real exponents, and does not seem to be related to Herglotz-like integrals. Assume $a,b\in\mathbb{R}$. Note that $$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln ... 11 Define:$$J_{{n}} \left( x,y,z \right) =\int _{-1}^{1}\!{\frac {\ln \left( xz+y \right) }{\sqrt {1-{x}^{2}} \left( xz+y \right) ^{n}}}{dx}\quad: n\in \mathbb{Z},\, 0\le z\le y \in \mathbb{R} \tag{1}$$and note that:$$\begin{aligned} {\frac {\partial }{\partial y}}J_{{n}} \left( x,y,z \right) &=\int _{-1 }^{1}\!{\frac {1}{\sqrt {1-{x}^{2}} \left( xz+y ...

8

Here is a proof of Cleo's answer. Rewrite the integral as \begin{align*} I &= \int_0^1 \frac{\log(1-x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &= \int_0^1 \frac{\log(1-x^2)-\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &= \int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \end{align*} The first integral can ...

7

This solution also appears on MathOverflow. We can think of $I_{n}$ as being a classical partition function for $n$ beads on a circle which cannot pass through each other, with logarithmic interaction potential between each bead and its next-to-nearest neighbors on either side. For $I_{2n}$ the beads fall into two colors" which do not have logarithmic ...

7

\begin{align*} I&=\int^1_0\frac{1-x^2+(1+x^2)\log x}{x+1}\frac{dx}{x\log^3x}\\ &=\left.\frac{-1}{2\log^2x}\frac{1-x^2+(1+x^2)\log x}{x+1}\right|^1_0-\int^1_0\frac{-1}{2\log^2x}\frac{\partial}{\partial x}\left(\frac{1-x^2+(1+x^2)\log x}{x+1}\right)dx\\ &=\int^1_0\frac{1}{2\log^2x}\frac{\partial}{\partial x}\left(1-x+\frac{(1+x^2)\log ... 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.$$We ... 5$$ \begin{align} \int_0^1\frac{x^3-x^2}{\log(x)}\mathrm{d}x &=\int_0^1\frac{x^4-x^3}{\log(x)}\frac{\mathrm{d}x}{x}\\ &=\lim_{a\to0}\int_a^1\frac{x^4-x^3}{\log(x)}\frac{\mathrm{d}x}{x}\\ &=\lim_{a\to0}\left(\int_a^1\frac{x^4-1}{\log(x)}\frac{\mathrm{d}x}{x} -\int_a^1\frac{x^3-1}{\log(x)}\frac{\mathrm{d}x}{x}\right)\\ ...

4


3

We start from the contour integral evaluated through Cauchy's theorem: $$\oint_{\Gamma_{\epsilon,R}}\mathrm{d}z\frac{z^2}{e^z+1} = 0$$ where $\Gamma_{\epsilon, R}$ is the rectangle of vertices $0,\ R,\ R+i2\pi,\ i2\pi$, indented at the singularity $i\pi$ with a semicircle of radius $\epsilon$. Splitting it into its natural branches yields: $$... 3 Yes, it's discontinuous, but you can still take the integral from -1 to 0, and 0 to 2. This is a particularly nice one to look at geometrically since you can directly calculate the area of these curves. So I'll just note that the magnitude of this curve is -1 for x < 0 and +1 for x > 0. So you have an area of -1*1 for the first section, and ... 3 This is a possible way (I assume a\geq 0, b\geq 0): 1) substitute x= a \sin t . 2) integrate the resulting integral by parts to get rid of \ln 3) you should end up with$$-\int_0^{\pi/2}\frac{2 a^2 b \sin^2 t\,dt }{b^2 -a^2 \cos^2 t} .$$4) it is possible to integrate the last integral by elementary means or using the residue theorem. You should ... 2 It is enough to compute \int \ln(a+\sqrt{1-x^2}) dx. Do a change of variables x=\sin t and integrate by parts:$$\int \ln (a+\cos t) \cos t \, dt = \ln(a+\cos t) \sin t + \int \frac{\sin^2 t}{a+\cos t} \, dt$$Use the substitution u = \tan t/2 so that \sin t = 2u/(1+u^2), \cos t = (1-u^2)/(1+u^2) and dt = 2/(1+u^2) du. This is known as the ... 2 Following my answer in this question, we have$$\int^1_0x^nK(\sqrt{x})K(\sqrt{1-x})dx\\ =\int^1_0(\frac\pi2\sum^{\infty}_{m=0}\frac{(2m)!^2}{2^{4m}(m!)^4}x^{m+n})K(\sqrt{1-x})dx\\ =\frac\pi2\sum^{\infty}_{m=0}\frac{(2m)!^2}{2^{4m}(m!)^4}\int^1_0x^{m+n}K(\sqrt{1-x})dx\\ ...

2

Substitute $x=\tan(\theta)$: \begin{align} \int_0^1\frac{\log(1+x^2)}{1+x^2}\,\mathrm{d}x &=2\int_0^{\pi/4}\log(\sec(\theta))\,\mathrm{d}\theta\\ &=-2\int_0^{\pi/4}\log(\cos(\theta))\,\mathrm{d}\theta\\ &=-\int_0^{\pi/4}\left[\log(1+e^{i2\theta})+\log(1+e^{-i2\theta})-2\log(2)\right]\,\mathrm{d}\theta\\ ... 2\int^∞_{-∞}\cos(\pi t) \,dt= \int^∞_{0}\cos(\pi t)\,dt + \int^0_{-∞}\cos(\pi t)\,dt$$Indeed, you are indeed correct that \int^0_{-\infty}cos(\pi t)\,dt diverges. And as you stated at the start, the integral is convergent if and only if split integrals are both convergent. Since you showed one of the two integrals in divergent, the entire ... 2 Let$$I_1=\int_{-1}^0{\frac{e^{\frac{x}{c}}}{\sqrt{-x}(x+2)}dx}$$and$$I_2=\int_{0}^\infty{\frac{e^{\frac{x}{c}}}{\sqrt{x}(x+2)}dx}$$and the given integral is I_1+I_2 and since$$\frac{e^{\frac{x}{c}}}{\sqrt{-x}(x+2)}\sim_0\frac1{2\sqrt{-x}}$$then I_1 exists, moreover we have$$\frac{e^{\frac{x}{c}}}{\sqrt{x}(x+2)}\sim_\infty ...

2

This integral is related to Fresnel S integral to which you arrive using the change of variable suggested by 7raisen7. For your problem, the antiderivative is $$\sqrt{\pi } S\left(\frac{2 \sqrt{x}}{\sqrt{\pi }}\right)$$ and the integral $$\sqrt{\pi } S\left(\sqrt{2}\right)=1.2654828001827241355...$$

2

Sub $x=a u$; then $$\int_0^a dx \, \log{x} \, \log{(a-x)} = a \int_0^1 du \left [\log^2{a} + \log{a} \left (\log{u} + \log{(1-u)}\right ) + \log{u} \, \log{(1-u)}\right ]$$ The first three integrals are straightforward; the middle two may be evaluated using the antiderivative $$\int dx \, \log{x} = x \log{x} - x +C$$ For the final integral, you can ...

1

This problem is awful (at least to me !). I was wondering if a CAS could do it; it did and I am sure that you will enjoy the result $$\frac{G_{3,3}^{3,2}\left(1\left| \begin{array}{c} 0,\frac{3}{4},1 \\ 0,0,0 \end{array} \right.\right)}{\Gamma \left(\frac{1}{4}\right)}$$ in which appears the Meijer G function (I am sure you are happy to know that !). The ...

1

Very accurate approximations can be computed thanks to series expansions such as the example given by Claude Leibovici (13 exact digits in case of $S(10)$) Other methods of numerical calculs leads to a lot of numerical appoximations of various kind. Some examples are compared below. Many surprising formulas are very easily obtained with the method of ...

1

Yes. The following general result helps to see why: Let $\mathcal{P}=\lbrace a=x_1<x_2<x_3<\cdots<x_n=b\rbrace$ be a partition; and $f$ a function defined on $[a,b]$ for any choice of $s_i\in [x_i,x_{i+1}]$ the sum $$\mathcal{S}=\sum_{i=1}^{n} f(s_i) \cdot (x_{i+1}-x_{i})$$ is a Riemann sum of $f$ associated to the partition $\mathcal{P}$. Let ...

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