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The integral being considered is, and is evaluated as, the following. \begin{align} I &amp;= \int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx \\ &amp;= \int_{0}^{\infty} \frac{dx}{1 + e^{...

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Consider the integral form of the Gamma function, \begin{align} \Gamma(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, dt \end{align} taking the derivative with respect to $x$ yields \begin{align} \Gamma'(...

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\begin{align} f(x) &amp;= \ln(2 + x^{2}) = \ln 2 + \ln\left( 1 + \frac{x^{2}}{2}\right) \\ &amp;= \ln 2 + \sum_{k=1}^{\infty} \frac{(-1)^{k-1} \, x^{2k}}{2^{k} \, k} \\ &amp;= \ln 2 + \frac{x^{2}}{2} -...

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Consider the integral \begin{align} I = \int_{0}^{1} \frac{\ln^{2}(x)}{1 - x + x^{2}} \ dx \end{align} Now consider the factorization of $x^{2} - x + 1$ which is $(x - a)(x-b)$ where $a$ and $b$ are ...

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Consider the integral \begin{align} I = \int_{0}^{\infty} \ln(1-e^{-ax}) \ \cos(bx) \ dx. \end{align} Expand the logarithm to obtain \begin{align} I &amp;= - \sum_{n=1}^{\infty} \frac{1}{n} \ \int_{0}^...

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In problems such as these one of the most efficient ways to calculate powers of a matrix is through diference equations. For the matrix of this problem the following is obtained. Given \begin{align} ...

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Consider the integral \begin{align} I = \int_{-\pi/2}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx. \end{align} This may also be seen as \begin{align} I &amp;= \int_{0}^{\pi/2} \frac{\sin^{...

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From Wolfram Gamma Function equations (35)-(37) provide \begin{align}\tag{1} \frac{1}{\Gamma(x)} = x + \gamma x^{2} + \sum_{k=3}^{\infty} a_{k} x^{k} \end{align} where, $a_{1}=1$, $a_{2}=\gamma$, \...

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The integral in question, \begin{align} I = \int_{0}^{1} \ln \left( \frac{1+ax}{1-ax} \right) \ \frac{dx}{x \sqrt{1-x^{2}}} \end{align} can be separated into the two integrals \begin{align} I = \int_{...

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Given \begin{align} I = \int_{a}^{\infty} \frac{e^{-x}}{\sqrt{x-a}} \, dx \end{align} let $t = x-a$ to obtain \begin{align} I &amp;= \int_{0}^{\infty} e^{-a - t} \, t^{(1/2) - 1} \, dt \\ &amp;= e^{-a}...

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For the integral \begin{align} I = \int_0^1 \frac{\operatorname{arctanh}\left(\sqrt{1-\frac{u}{2}}\right)\sqrt{\frac{2 \pi \sqrt{1-u}}{u-2}+\pi } }{u\sqrt{1-u}} \, du \end{align} let $t^{2} = 1-u$ to ...

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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^{-...

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Consider the integral \begin{align} I = \int_{0}^{1} \frac{K(x)}{\sqrt{1-x^2}} \ dx \end{align} where $K(x)$ is the complete elliptic integral of the first kind. It can be shown that the ...

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The Chebyshev polynomials of the first kind are given by \begin{align} T_{n}(x) &amp;= \frac{n}{2} \sum_{k=0}^{[n/2]} \frac{(-1)^{k}}{n-k} \binom{n-k}{k} (2x)^{n-2k} \\ &amp;= \frac{1}{2} \left[ (x - \...

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\begin{align} \int_{0}^{\infty} (x - \ln(e^{x} - x)) \ \mathrm{d}x &amp;= \int_{0}^{\infty} (x - \ln(e^{x}) - \ln(1-x e^{-x}) ) \ \mathrm{d}x \\ &amp;= - \int_{0}^{\infty} \ln(1 - x e^{-x}) \ \mathrm{...

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To save from typing a result a highlight will be listed for now. The integral in question is a reduction of the more general integral \begin{align} \int_{0}^{\infty} \frac{J_{\mu}(at) \, J_{\nu}(bt)}...

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Neumann's addition theorem is given by \begin{align} J_{0}\left(\sqrt{x^{2} + y^{2} - 2 x y \cos\phi}\ \right) = J_{0}(x) J_{0}(y) + 2 \sum_{n=1}^{\infty} J_{n}(x) J_{n}(y) \cos(n\phi). \end{align} ...

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Process 1: It is known that \begin{align} \int_{0}^{1} x^{\nu -1} (1+x)^{\lambda} (1-x)^{\mu -1} \, dx = B(\mu, \nu) \, {}_{2}F_{1}(- \lambda, \nu; \mu+\nu; -1) \end{align} for which \begin{align} \...

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By making use of the integral $$\int_{0}^{1} \frac{(1-x)^2}{1-x^3} \, dx = \frac{1}{2} \, \left(\frac{\pi}{\sqrt{3}} - \ln 3 \right)$$ one can take the following path. \begin{align} S &amp;= \sum_{k=0}...

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Since $(\sqrt{3})^{4} = 9$ then $$\log_{9}\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{4} \, \log_{9}\left(\frac{1}{9}\right) = - \frac{1}{4} \, \log_{9}(9) = - \frac{1}{4}.$$

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It is evident that $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right) = \frac{\pi^2}{12} - \ln^{2}2$$ and can be evaluated by following the pattern: Consider the ...

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For the integral $$\int x^{x^2+1} \, (2\ln x+1) \, dx$$ let $u = x^{2} \, \ln x$ for which $du = (2x \, \ln x + x ) \, dx$ and \begin{align} I &amp;= \int x^{x^{2} + 1} \, (2 \, \ln x + 1) \, dx \\ &...

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Starting with the Beta function \begin{align} B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, dt \end{align} differentiate with respect to $x$ and $y$ twice. This leads to \begin{align} I &amp;= \int_{0}...

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From here the relation \begin{align} \sum_{n=1}^{\infty} \zeta(n+1) \ x^{n} = - \gamma - \psi(1-x) \end{align} can be obtained. By letting $x$ go to $-x$ and adding the result the following is ...

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Consider the integral \begin{align} I = \int_{0}^{1} \frac{\ln\left( \frac{1+x}{1-x} \right)}{x \sqrt{1-x^{2}}} \ dx \end{align} when the transformation $x = \tanh(t)$ is made. The resulting integral ...

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The method used here is that of the generating function. Let $S_{n}$ be the series to be summed \begin{align} S_{n} = \sum_{k=0}^{n} \binom{n+k}{2k} \ 2^{n-k}. \end{align} The generating function and ...

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Although this question is older, the question and result are vague. The following may be of assistance. For the integral $$\int W(a x) \, dx$$ let $t = W(a x)$, or $x = \frac{1}{a} \, t \, e^{t}$ ...

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There is an alternate method and is as follows. Notice that $$\frac{1}{n(n+1)(n+2)} = \frac{(n-1)!}{(n+2)!} = \frac{1}{2!} \, B(n,3)$$ where $B(x,y)$ is the Beta function. Using an integral form ...

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The differential set $$(1+y^{2}) \,dx = (1+x^{2})\, dy$$ can be integrated as seen by \begin{align} \int \frac{dy}{1+y^{2}} = \int \frac{dx}{1+x^{2}} \end{align} and leads to \begin{align} \tan^{-...