# Tag Info

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### Integral of the product of a Gaussian and a logarithmic function of Laguerre polynomials

It took some effort, but I found a closed expression. First, we use the close form and roots of the Laguerre polynomials to express the integral as: f_n=\frac{1}{2}\sum_{q=0}^n\sum_{\...
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### Doubt in Integral of $\int_{0}^{\frac{\pi}{2}}\frac{x}{\sin^3x + \cos^3x} \,\mathrm{d}x$

The standard "brute force" procedure once you reach $$I = \frac{\pi}{4}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin^3x + \cos^3x} \,\mathrm{d}x$$ is to recognize you have a rational function of ...
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### Evaluate a hard integral: $\int_{t'=0}^{\infty} (\alpha+\beta t')^{\gamma-1}\left(1-e^{-k(t'-t_{0})}\right)e^{-(r-g)(t'-t_{0})} \, \mathrm{d}t'$

For my ease of mind, I will start by letting $\newcommand{\r}{\rho} \newcommand{\a}{\alpha} \newcommand{\b}{\beta} \newcommand{\t}{\tau} \newcommand{\k}{\kappa} \r := r-g$ and $\t := t'-t_0$ to give ...
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### Find $t$ such that $\int_0^t x^x dx = 1$

I don't think that there is any analytical solution. For a numerical one, you can solve by Runge-Kutta the ODE $$\frac{dt}{dI}=t^{-t}$$ for $I\in[0,1]$. This avoids having to combine a solver and an ...
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### Tricky integral $\int_0^\pi{12\cos x\ \mathrm{sech}(\frac \pi2 \tan\frac x2)}\mathrm{d}x=\pi^2$

We will prove $$\int_{0}^{\pi}\cos\left(x\right)\operatorname{sech}\left(\frac{\pi}{2}\tan\frac{x}{2}\right)\,dx=\frac{\pi^{2}}{12}\,.$$ PROOF Let $\mathcal{I}$ be the integral in question. We use ...
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### Find $t$ such that $\int_0^t x^x dx = 1$

$$\int_0^t x^x dx = 1$$ As far as I know there is no closed form for $t$. A non-standardized closed form could be defined from the inverse of the Sphd$(1,t)$ function. But this would be purely ...
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### Evaluate Integral : $\int^1_0\frac{\ln(x)}{4\pi^2+\ln^2(x)}\cdot \frac{1}{1-x}dx$

Let $I$ denote the integral. Then \begin{align*} I &= - \int_{0}^{\infty} \frac{t}{4\pi^2 + t^2} \frac{e^{-t}}{1 - e^{-t}} \, \mathrm{d}t \tag{$x = e^{-t}$} \\ &= - \sum_{n=1}^{\infty} \int_{0}...
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### How to show $\lim\limits_{a\rightarrow 1^-}\frac{\int_{0}^{a}\frac{x^{2}}{1-x}\,\mathrm{d}x}{\int_{0}^{a}\frac{x}{1-x}\,\mathrm{d}x}=1$?

Let $f(x)$ and $g(x)$ be continuous on $[0,1)$ and $\displaystyle\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}g(x)=+\infty.$ If the limit $\displaystyle\lim_{x\to 1^-}{f(x)\over g(x)}$ exists then by the l'...

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### How to Evaluate $\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} .$?

by putting $u=\frac{x}{\sqrt{x^2+1}}$ we have $$I=\int_0^1 \frac{u\sin^{-1}u}{u^4+2u^2+13} du=\int_0^{\infty} \frac{x\tan^{-1}x}{16x^4+28x^2+13} dx$$ where $\sin^{-1}u=\tan^{-1}x$ and now by using ...
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