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$\int_1^2\dfrac{dx}{1+x+\ln x}$ $=\int_0^{\ln2}\dfrac{d(e^x)}{1+e^x+x}$ $=\int_0^{\ln2}\dfrac{e^x}{e^x+x+1}dx$ $=\int_0^{\ln2}\dfrac{1}{1+(x+1)e^{-x}}dx$ $=\int_0^{\ln2}\left(1+\sum\limits_{n=1}^\infty(-1)^n(x+1)^ne^{-nx}\right)dx$ $=\left[x+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nn!(x+1)^ke^{-nx}}{n^{n-k+1}k!}\right]_0^{\ln2}$

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According to http://upload.wikimedia.org/wikipedia/en/math/7/2/a/72a1058ad2087aec467af24bddcf9479.png, since every convergent definite integral can be expressed as the infinite sum, perhaps this is the simplest explanation.

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$$\int^∞_{-∞}\cos(\pi t) \,dt= \int^∞_{0}\cos(\pi t)\,dt + \int^0_{-∞}\cos(\pi t)\,dt$$ 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 ...

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The basic idea behind this is to first observe that $$\overline{f(e^{i\theta})} = f(\overline{e^{i\theta}}) = f(e^{-i\theta}).$$ Then we note that the product is $$f(e^{i\theta}) \overline{f(e^{i\theta})} = \sum_{k=0}^n c_k e^{k i\theta} \sum_{m=0}^n c_m e^{-m i \theta} = \sum_{k=0}^n \sum_{m=0}^n c_k c_m e^{(k-m)i\theta}.$$ Then consider integrating a ...

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HINT: substitute $\ln{x}=t$ Denominator gives an $x$, so $\frac{1}{x}dx=dt$ U then have $\int\frac{t^2}{t-2}dt-\int\frac{3t}{t-2}dt+\int\frac{3}{t-2}dt$ which can be done

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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 ...

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Just for your information, I used a CAS without any success for the general case. However, I obtained some formulas. For $p=2$, $$\frac{1}{8} \left(-\gamma _1\left(\frac{1}{8}\right)+\gamma _1\left(\frac{5}{8}\right)-\sqrt{2} (\gamma +\log (8)) \left(\pi +2 \log \left(\cot \left(\frac{\pi }{8}\right)\right)\right)\right)$$ For $p=3$, $$\frac{1}{36} ... 3 By the way,$$ \sum_{n=1}^\infty \frac{1}{n^n} \approx 1+\frac{\sin \left(\frac{\pi }{e}\right)}{\pi } $$seems to be a better approximation. According to RIES,$$\sum_{n=1}^\infty \frac{1}{n^n} \approx\left(\frac{\phi +\pi }{2}\right)^{\frac{1}{\pi+\frac{1}{4} }}$$is still better but not as good as the one proposed by JJacquelin. Next one, please ! 6 This solution also appears on MathOverflow. We can think of I_{n} as being a classical partition function for n beads on a circle which cannot pass through each other, with logarithmic interaction potential between each bead and its next-to-nearest neighbors on either side. For I_{2n} the beads fall into two colors" which do not have logarithmic ... 2$$I=\frac{\Gamma\left(\frac14\right)^2}{4\,\sqrt{2\,\pi}}\Big(\ln2-\pi\Big).$$0 Case 1: k=0 , c\in\mathbb{R^-} , b\in\mathbb{R^+} Then \int_b^\infty\dfrac{e^{ct}t}{\sqrt{t^2-b^2}}dt =\int_0^\infty\dfrac{e^{cb\cosh t}b\cosh t}{\sqrt{(b\cosh t)^2-b^2}}d(b\cosh t) =\int_0^\infty\dfrac{b^2e^{bc\cosh t}\sinh t\cosh t}{b\sinh t}dt =\int_0^\infty be^{bc\cosh t}\cosh t~dt =bK_1(-bc) Case 2: c=0 , k\in\mathbb{R^+} , ... 1 The curves need to be thought of as with respect to the line y=1. What's the distance between the point (x,\sqrt{x}) and the line y=1? What's the distance between (x,x) and y=1? Now compute$$A=\int_0^1 \pi\left((1-x)^2-(1-\sqrt{x})^2\right)dx.$$4 In general, if a_n>0, a_n\to 0 and a_n decreasing then the series$$ \sum_{n=1}^\infty (-1)^{n-1} a_n, $$converges (not necessarily absolutely), say to s, and$$ s_{2n-1}<s_{2n+1}<s<s_{2n+2}<s_{2n}, $$where s_n=\sum_{k=1}^n (-1)^{k-1}a_k. In particular,$$ s_n-s\quad\text{and}\quad s_{n+1}-s, $$have different signs. This implies ... 0 Mathematica: Assuming[{ta > 0, ta > te, te > 0}, Block[{f, igrand, lamtbl}, f[ta_, lambda_] := lambda/(4*(Sqrt[Pi ta^3])) Exp[-lambda^2/(4 ta)]; igrand[ta_, te_, lambda_] := f[ta, lambda] Erf[lambda/(2*Sqrt[ta - te])]; lamtbl = {0.4, 0.8, 1.2, 2, 2.4}; Table[ NIntegrate[igrand[ta, te, lamtbl[[which]]], {ta, te, Infinity}], ... 0 There is a closed form expression for this integral and I am sure you will enjoy it$$\frac{3}{2} \pi \left(2 \text{Hypergeometric2F1}^{(0,0,1,0)}\left(\frac{3}{4},\frac{3}{4},1,-8\right)+\text {Hypergeometric2F1}^{(0,1,0,0)}\left(\frac{3}{4},\frac{3}{4},1,-8\right)+\text {Hypergeometric2F1}^{(1,0,0,0)}\left(\frac{3}{4},\frac{3}{4},1,-8\right)-\, ...

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The bounds are $\bigg[1,\dfrac\pi2\bigg]$, since $x^{2n}\in[0,x^2]$ for $x\in[0,1]$ with $n>1$.

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You will have to draw the region.Region in xy plane is bounded by $x=1-(y-1)^2$ , a parabola and y=1. and from there z from 0 to z=x-2, a plane. Now if you project that volume on yz plane you will see its a rectangle bounded by z=2 and y=1, Now we can setup this integral, in order dxdydz $$\int_0^2\int_0^{1}\int_0^{1-(y-1)^2}f(x,y,z)dxdydz$$

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Though this isn't an answer, this is interesting enough to me but too large for a comment. Based on Vladimir's solution, if we know $$f(a) = \int_0^\infty \frac{\ln[(1+x^a)/2]}{\ln x} \frac{1}{1+x^2}dx = a\frac{\pi}{4}$$ then we should have $$f'(a) = \int_0^\infty \frac{\ln x e^{a\ln x}}{\ln x (1+e^{a \ln x})} \frac{1}{1+x^2} dx = \pi/4$$ or $$f'(a) = ... 11 This integral can be evaluated in a closed form for arbitrary real exponents, and does not seem to be related to Herglotz-like integrals. Assume a,b\in\mathbb{R}. Note that$$\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 ...

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Hint: $\int_b^c\dfrac{\sqrt{x}e^x\text{erfc}(\sqrt{x})}{\sqrt{a-x}}dx$ $=\int_b^c\dfrac{\sqrt{x}e^x}{\sqrt{a-x}}dx-\int_b^c\dfrac{\sqrt{x}e^x\text{erf}(\sqrt{x})}{\sqrt{a-x}}dx$ $=\int_b^c\sum\limits_{n=0}^\infty\dfrac{x^{n+\frac{1}{2}}}{n!\sqrt{a-x}}dx-\int_b^c\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}n!x^{n+1}}{(2n+1)!\sqrt\pi\sqrt{a-x}}dx$ (according to ...

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In my opinion, the major problem is with the absolute value. However, the integrand cancels at $\sqrt{\frac{\pi }{2}}$; integrating from $0$ to this value gives as a result $$\sqrt{\frac{\pi }{2}} C(1)$$ which is slightly smaller than $1$ ($0.977451$) which is the target value. So, still forgetting the absolute value and computing the integral from ...

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With $y^4=x+1$, Maple gets the antiderivative $$V(y)=2\,\ln \left( y+i \right) \pi +2\,\ln \left( y-i \right) \pi -4\, \ln \left( 1+y \right) \ln \left( y \right) -4\,i{\rm Li}_2 \left( 1-iy \right) +4\,i{\rm Li}_2 \left( 1+iy \right) -4\,{ \rm Li}_2 \left( -y \right) -4\,{\rm Li}_2 \left( 1-y \right)$$ Now $V(0)=-8 \;\mathrm{Catalan} + \pi^2/3$ and ...

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\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)\\ ... 1 If f(a)\neq f(b), say f(a) is larger. Then we can always shift the interval towards the direction of a to make it shorter (since f is continuous). 3I(n)=\int_0^\infty(x+1)^n\frac{dx}x\iff\int_0^{\infty}\frac{\ln(x+1)}{x\cdot(x+1)^\frac14} dx=I'\bigg(-\frac14\bigg)$$Let t=\dfrac1{x+1} , and recognize the expression of the beta function in the new integral, then use the reflection formula, but do not compute its value, since it will be divergent ! Rather, derive it directly with regard to n. The ... 14 Sub x=e^{-u}, dx = -e^{-u} du. Then the integral is$$\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}}$$The change in the order of integration is justified by Fubini's ... 1 This problem is awful (at least to me !). I was wondering if a CAS could do it; it did and I am sure that you will enjoy the result$$\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)}$$in which appears the Meijer G function (I am sure you are happy to know that !). The ... 5 When 0<x<1, we have the identity K(x^{-1})=x(K(x)-iK(\sqrt{1-x^2})). Therefore$$\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.$$We ... 0 The function you have to integrate is not defined at 0, but it is bounded on (0,\frac{\pi}{2}], so this is a regular Riemann Integral, not an improper one. For 0<h<\frac{\pi}{2} the function is continuous on [h,\frac{\pi}{2}], so you can apply the Fundamental Theorem of Calculus, and then take the limit for h\rightarrow 0^+ of the function ... 0 Your integral is an improper integral because \cot x is not continuous at x=0. To correct this issue, you should set up the problem as$$ \int_0^{\pi/2}x\cot x\,dx=\lim_{a\to 0^+}\int_a^{\pi/2}x\cot x\,dx $$Then evaluate the integral$$ \int_a^{\pi/2}x\cot x\,dx $$This will give you an expression in a. Once you have this expression take a\to 0. 0 I do not know how to answer your question. However, in order you to challenge your challenger, I give you a few amazing results for$$f(n)=\int_0^\infty\frac{\log\left(\frac{1+x^a}{1+x^b}\right)}{\left(1+x^2\right)\log x}dx$$in which a=2n+\sqrt{4 n^2-1} and b=n+\sqrt{n^2-1}.$$f(1)=\frac{1}{4} \left(1+\sqrt{3}\right) \pif(2)=\frac{1}{4} ...

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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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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: $$... 1 The region is a solid one which is bounded by the plane z=0,z=2-x and a cylinder x=1-(y-1)^2 and 0\le y\le 1: This can be shown by the solid red region below: 1 I_n(x)=\int_0^x\cos^n2t~dt =\int_0^x\cos^{n-1}2t\cos2t~dt =\dfrac{1}{2}\int_0^x\cos^{n-1}2t~d(\sin2t) =\left[\dfrac{1}{2}\sin2t\cos^{n-1}2t\right]_0^x-\dfrac{1}{2}\int_0^x\sin2t~d(\cos^{n-1}2t) =\dfrac{1}{2}\sin2x\cos^{n-1}2x+(n-1)\int_0^x\cos^{n-2}2t\sin^22t~dt =\dfrac{1}{2}\sin2x\cos^{n-1}2x+(n-1)\int_0^x\cos^{n-2}2t(1-\cos^22t)~dt ... 2 Let$$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 ...

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