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6

As an alternative, we may apply the substitution $x\mapsto\sinh^2{x}$. \begin{align} \int^1_0\frac{\ln{x}}{\sqrt{x(x+1)}}{\rm d}x &=4\int^{-\ln(\sqrt{2}-1)}_0\ln(\sinh{x})\ {\rm d}x\\ &=4\int^{-\ln(\sqrt{2}-1)}_0\left(\color{green}{\ln\left(1-e^{-2x}\right)}-\ln{2}+x\right)\ {\rm d}x\\ ...


5

By making the Euler substitution $\sqrt{x^{2}+x} = x+t$, we find $$ \begin{align} \int_{0}^{1} \frac{\log(x)}{\sqrt{x^{2}+x}} \, dx &= 2 \int_{0}^{\sqrt{2}-1} \frac{\log \left(\frac{t^{2}}{1-2t}\right)}{1-2t} \, dt \\ &= 4 \int_{0}^{\sqrt{2}-1} \frac{\log t}{1-2t} \, dt - 2\int_{0}^{\sqrt{2}-1} \frac{\log(1-2t)}{1-2t} \, dt. \end{align}$$ The first ...


5

Let: $$ f(s)=\int_{0}^{1}\frac{x^s}{\sqrt{1+x}}\,dx. $$ We have, by integration by parts: $$ f(s+1)+f(s) = \int_{0}^{1} x^s\sqrt{1+x}\,dx = \frac{1}{s+1}\left(\sqrt{2}-\frac{1}{2}\,f(s+1)\right)$$ hence: $$ \left(2s+3\right)\, f(s+1)+(2s+2)\,f(s) = 2\sqrt{2},$$ $f(0)=2\sqrt{2}-1$ and $\lim_{s\to +\infty}f(s)=0$. We have: $$ f(s)=\sum_{n\geq 0}\frac{(-1)^n ...


3

I tried 2 ways to find a closed form, though unsuccessful up to this point. 1st trial. Let $I$ denote the integral, and write $$ I = 8 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \int_{0}^{\infty} \frac{e^{-x}(1 - e^{-(2n+1)x})}{x (1 + e^{-x})^{2}} \, dx. $$ In order to evaluate the integral inside the summation, we introduce new functions $I(s)$ and ...


3

We can write $$\frac{1}{1+a^n}=\left[\frac{\partial}{\partial t}\ln\left(1+ta^{-n}\right)\right]_{t=1},$$ which implies that $$\sum_{n=0}^{\infty}\frac{1}{1+a^n}=\left[\frac{\partial}{\partial t}\ln\prod_{n=0}^{\infty}\left(1+ta^{-n}\right)\right]_{t=1}= \left[\frac{\partial}{\partial t}\,\ln\left(-t;a^{-1}\right)_{\infty}\right]_{t=1},$$ where ...


2

By expanding $\frac{1}{1+a^n}$ as a geometric series, $$\sum_{n\geq 1}\frac{1}{1+a^n}=\sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{a^{mn}}=\sum_{l\geq 1}\frac{1}{a^l}\sum_{d\mid l}(-1)^{\frac{l}{d}+1}=\sum_{l\geq 1}\frac{g(l)}{a^l}$$ where $g(l)$ is an arithmetic function counting the difference between the number of odd divisors and the number of even ...


2

Let's make a substitution $\sin(\theta)=y$ $$ I(x,z)=\int_{0}^{\sin(z)}\frac{J^2_{1}(x y)}{y^2\sqrt{1-y^2}}dy $$ Because $z\approx0.1$ we may that assume $\sin(z)\approx z$. Furthermore we perform another subsitution $\frac{y}{z}=q$ yielding $$ I(x,z)=z\int_{0}^{1}\frac{J^2_{1}(x z q)}{(zq)^2\sqrt{1-(zq)^2}}dq $$ As a first step to simplify the problem we ...


2

The elements in $a$ must be non-negative. (Take $a=-1$ for a start of a counterexample) If $a\ge 0$ then $$ a^Tb_i = \sum_{j=1}^d a_j b_{i,j} \le \sum_{j=1}^d a_j \max_i b_{i,j}= a^T (\max b_i). $$ Now take the maximum on the left-hand side.


1

Your function is called the Generalized Exponential Integral $E_p(x)$ and is described in http://dlmf.nist.gov/8.19: $$E_p(x) = x^{p-1}\int_x^\infty \frac{e^{-t}}{t^p}\: d t \; = \int_1^\infty \frac{e^{-xt}}{t^p}\: d t\; , $$ Without a specific reference to the actual open source code, I guess that the discrete recurrence formula based on $E_1$ or your ...


1

The op $D^{-1}D_{h,x}$ and its inverse $DD_{h,x}^{-1}$ are classic operators associated with the Bernoulli numbers (and polynomials) and the Euler-Maclaurin formula.


1

I think it is a correct answer: $$e^{hD}f = \sum \frac{h^n}{n!}D^n f = f(x+h)$$ so $$D_{h,x}f = \frac{f(x+h)-f(x)}{h}$$ Given $$g(x) = \frac{f(x+h)-f(x)}{h}$$ we may reconstruct $f(x)$ as $$f(x) = h\sum_{n=1} g(x - nh)$$ Now $g(x - nh) = e^{-nhD}g$, that is $$f = h\sum e^{-nhD}g = h\sum (e^{-hD})^n g= h(\frac{1}{1 - e^{-hD}} - 1)g = \frac{h}{e^{hD} - 1}g$$ ...


1

If $\mathcal{I} = D^{-1}$ then wouldn't $$ D^{-1}_{h,x}[f] = \mathcal{I} \left[ \frac{\ln(fh+1)}{h} \right], $$ because then $$ D^{-1}_{h,x}\left[D_{h,x}[f]\right] = f? $$


1

Assuming that you are talking only about $\Bbb R\to\Bbb R$ functions, for a such function, this equation holds: $$f(x)=x\;\forall x\in f(\Bbb R)$$ So, for any nonempty $A\subset \Bbb R$ define $f(x)=x$ for $x\in A$, and $f(x)$ to be any $y\in A$ otherwise. Any such function will hold $f\circ f=f$. The sets $A$ correspondig to your examples are, ...


1

Hint. You have an Euler-type integral, you may then rewrite your initial integral in terms of the Appel hypergeometric function $$ F_1(a,b_1,b_2,c;x,y)=\dfrac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1\frac{v^{a-1}(1-v)^{c-a-1}}{(1-xv)^{-b_1}(1-yv)^{-b_2}}~dv $$ where $\Re c>\Re a>0$ with the change of variable $$ v=\frac{u}{1-e^{-\lambda x}}. $$


1

$y=\phantom{}_2 F_1(a,b;c;z)$ is the regular solution of the ODE: $$z(1-z)y''+\left[c-(a+b+1)z\right]y'-ab y=0 $$ hence $y=\phantom{}_2 F_1(-n,n;1/2;z)$ is the regular solution of the ODE: $$z(1-z)y''+\left(\frac{1}{2}-z\right)y'+ n^2 y=0 $$ and $y=\phantom{}_2 F_1\left(-n,n;\frac{1}{2};\frac{1-z}{2}\right)$ is the regular solution of the ODE: ...



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