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0

Let me answer my own question: I contacted the author and he was very kind to review my comments and agreed that there is indeed an error in the paper. The problem seems to originate in formulae $(1.6), (1.8)$ that are not valid for all fractional values of $u$. In the beginning of the section $2$ the author writes We now boldly assume that $(1.6)$ ...


1

For $I_1$ Notice: $p(z)=1+2z^2+2z^4=2\Big(z^2+az+b\Big)\Big(z^2-az+b\Big)$ where $a=\sqrt{\sqrt{2}-1}$ and $b=\dfrac{\sqrt{2}}{2}$ Therefore: $f(z)=\displaystyle \dfrac{2z^2}{1+2z^2+2z^4}=\dfrac{z}{2a\Big(z^2-az+b\Big)}-\dfrac{z}{2a\Big(z^2+az+b\Big)}$ ...


2

You forgot to replace the $dt$ with the proper expression in the substitution. With $t = \tan x$, we have $dt = \tan' x\,dx = (1+\tan^2 x)\,dx$, so the integral becomes $$\int_0^{\arctan x} \frac{1+\tan^2 x}{1+\tan^2 x}\,dx = \int_0^{\arctan x} 1\,dx = \arctan x$$ and analogously for the other integral.


4

$$ \begin{align} \int_0^\infty\frac{\log(x)}{x^n-1}\mathrm{d}x &=\int_{-\infty}^\infty\frac{x}{e^{nx}-1}e^x\,\mathrm{d}x\\ &=\int_0^\infty\frac{x}{e^{nx}-1}e^x\,\mathrm{d}x+\int_0^\infty\frac{x}{1-e^{-nx}}e^{-x}\,\mathrm{d}x\\ &=\int_0^\infty x(e^{(1-n)x}+e^{(1-2n)x}+e^{(1-3n)x}+\dots)\,\mathrm{d}x\\ &+\int_0^\infty ...


4

We can use a contour integral in the complex plane, as I showed here for the case $n=3$. Now, however, we use $$\oint_C dz \frac{\log^2{z}}{z^n-1}$$ where $C$ is the modified keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$. Let's evaluate this integral over the contours. As before, there are $8$ pieces to ...


1

In several places, we will use $\frac1{k(k+1)}=\frac1k-\frac1{k+1}$. $$ \begin{align} \int_0^1\sum_{k=1}^\infty\frac{x^k}{k^p}\mathrm{d}x &=\color{#C00000}{\sum_{k=1}^\infty\frac1{k^p(k+1)}}\\ &=\sum_{k=1}^\infty\left(\frac1{k^p}-\frac1{k^{p-1}(k+1)}\right)\\ &=\color{#C00000}{\zeta(p)-\sum_{k=1}^\infty\frac1{k^{p-1}(k+1)}}\\ \end{align} $$ ...


0

(as per David H's suggestion to use Integration By Parts) Hypothesis: The two definite integrals $D_1$ and $D_2$ have the same value, i.e. $D_1 = D_2$. Using integration by parts $$ \int uv' \,d\theta= uv-\int u'v \,d\theta $$ let us define $$ u = \frac{1}{(1-a\cos\theta)^3} \qquad v' = \cos\theta $$ So ($u'$ obtained from WolframAlpha) $$ ...


0

Hello there, this is Cleo who is using Anastasiya's M.S.E. account (>‿◠)✌ Does the integral below have a closed-form? \begin{equation} I=\int_{0}^{\Large\frac{\pi}{2}} \arctan \left( \frac{\cos x}{\sin x - 1 - \sqrt{2}} \right) \tan x \,\,dx \end{equation} ${\sf\mbox{Yes, it does.}}$ Here is the bonus answer for your question: where $\mathbf{G}$ ...


2

First, assume $\displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{n^2}=\dfrac{\pi^2}{6}$ Then: $\displaystyle \dfrac{\pi^2}{6}=\sum_{n=1}^{+\infty} \dfrac{1}{n^2}=\sum_{n=1}^{+\infty} \dfrac{1}{(2n)^2}+\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}=\dfrac{\pi^2}{24}+\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}$ Therefore: $\displaystyle \sum_{n=0}^{+\infty} ...


0

Let $\Phi(t) = \int_{-\infty}^t e^{-x^2/2} \, dx$. Then $$ \int_k^{\infty} x^2 e^{-x^2/2} \,dx = -\frac{d}{d\alpha}\Big|_{1/2}\int_k^{\infty}e^{-\alpha x^2}\,dx = - \frac{d}{d\alpha}\Big|_{1/2} \int_k^{\infty} e^{-(\sqrt{2 \alpha}x)^2/2}\,dx =\\ -\frac{d}{d\alpha}\Big|_{1/2} \Big( \frac{1}{\sqrt{2\alpha}} \int_{\sqrt{2 \alpha} k}^{\infty} e^{-x^2/2} \,dx ...


0

Put $$t^2=\frac{(x-\mu)^2}{2a^2} (\implies x=\sqrt 2\,a\,t+\mu)\implies dx=\sqrt2\,a\;dt$$ and then we get the integral $$\sqrt2\,a\int\limits_{\frac\mu{\sqrt2\,a}}^\infty (\sqrt{2t}\,a+\mu)^2e^{-t^2}dt$$ and playing around with this you'll get a part with the err function and another with the whole integral.


3

\begin{align} \int_0^{\Large\frac{\pi}{4}} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx&=\int_0^1\frac{\sqrt{y(1-y)}}{1+y^2}\,dy\quad\Rightarrow\quad y=\tan x\\ &=\int_0^\infty\frac{\sqrt{t}}{(1+t)(1+2t+2t^2)}\,dt\quad\Rightarrow\quad t=\frac{y}{1-y}\\ &=\int_0^\infty\frac{2z^2}{(1+z^2)(1+2z^2+2z^4)}\,dz\quad\Rightarrow\quad z^2=t\\ ...


2

The result happens to coincide with the conjectured form: $$\mathcal I=\left(\sqrt[4]{2}\,\cos\frac{\pi}{8}-1\right)\pi.$$ Derivation: make the change of variables $t=\tan x$. This transforms the integral into $$\mathcal I=\int_0^1\frac{\sqrt{t(1-t)}}{1+t^2}dt$$ Now since we have a mix of rational function with only square roots of a quadratic ...


1

I have got a series expansion, though do not know if it of any use getting a closed form expression. Substituting $z=\tan x$ $$I=\int_{0}^1 \dfrac{z^{1/2}(1-z)^{1/2}}{1+z^2}dz=\sum_{k=0}^\infty (-1)^k\int_{0}^1z^{1/2+2k}(1-z)^{1/2}dz\\=\sum_{k=0}^{\infty}(-1)^k\beta\left(2k+3/2,3/2\right)=\dfrac{\sqrt{\pi}}{2}\sum_{k=0}^{\infty}(-1)^k ...


1

That's just a start, but using the change of variable: $$ u = \sqrt{\tan(x)}\quad\Rightarrow\quad\mathrm du = \frac{1+u^4}{2u}\mathrm dx, $$ you get: $$ I= 2\int_0^1 \frac{u^2\sqrt{1-u^2}}{1+u^4}\mathrm du $$ Now, let: $u= \sin(t)\Rightarrow\mathrm du = \cos(t)\mathrm dt$ $$ I= 2\int_0^{\frac{\pi}{2}} \frac{\sin^2(t)\cos^2(t)}{1+\sin^4(t)}\mathrm dt $$ ...


3

I found it now, but it isn't very elegant: $$ \int_0^1 \ln\left(K(x)\right)\space dx =\int_0^1 \ln\left(x^{-x}\cdot(2\pi)^{-\frac{x}{2}}\cdot e^{\frac{1}{2}x(x+1)-\frac{\gamma}{2}x^2}\cdot\prod_{k=1}^\infty \left[\frac{e^{x+\frac{x^2}{2k}}}{\left(1+\frac{x}{k}\right)^{x+k}}\right]\right)\space dx ={-\int_0^1 {x\ln\left(x\right)\space dx}}-\int_0^1 \frac x ...


5

In general, an integral representation of the beta function is $$B(x,y) = \int_{0}^{1} \frac{t^{x-1} + t^{y-1}}{(1+t)^{x+y}} \ dt \ , \ \text{Re} (x), \text{Re}(y) >0 . $$ This can be derived by expressing the more well-known integral representation $$ B(x,y) = \int_{0}^{\infty}\frac{t^{x-1}}{(1+t)^{x+y}} \ dt \tag{1}$$ as $$\begin{align} B(x,y) &= ...


5

$$\begin{align*} \int_{0}^{1}\frac{1+x^{a}}{\left(1+x\right)^{a+2}}dx &=\int_{0}^{1}\frac{1}{\left(1+x\right)^{a+2}}dx+\int_{0}^{1}\frac{x^{a}}{\left(1+x\right)^{a+2}}dx \\ &=\left.-\frac{1+x}{\left(a+1\right)\left(1+x\right)^{a+2}}\right|_{0}^{1}+\left.\frac{\left(1+x\right)x^{a+1}}{\left(a+1\right)\left(1+x\right)^{a+2}}\right|_{0}^{1} \\ ...


3

First: $~\displaystyle 2\int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx=\int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx\qquad(1)$ Proof: Let $I=\displaystyle \int_0^{\tfrac{\pi}{12}} \log(\tan(3x))dx$ $\tan(3x)=\tan(x)\tan\big(\dfrac{\pi}{3}-x\big)\tan\big(\dfrac{\pi}{3}+x\big)$ $\displaystyle I= \int_0^{\tfrac{\pi}{12}} \log(\tan(x))dx+\int_0^{\tfrac{\pi}{12}} ...


1

For the variance one has $$Var(X\pm Y) = Var(X) + Var(Y) \pm Cov(X,Y)$$ so if $X$ and $Y$ are uncorrelated, i.e. $Cov(X,Y)=0$ (which does not mean independent, but if they are independent then they are uncorrelated) then $$Var(X-Y) = Var(X)+Var(Y)$$ and hence $$E[(X-Y)^2] = Var(X)+Var(Y) + E[X-Y]^2$$ if you look at wikipedia, having $X\sim ...


2

Let us assume that $$\Gamma\left(k+\frac{1}{2}\right)=2\int^\infty_0 e^{-x^2}x^{2k}\,dx=\frac{\sqrt{\pi}(2k)!}{4^k k!}$$ 1- for $k=0$ we have $$\Gamma\left(\frac{1}{2}\right)=2\int^\infty_0 e^{-x^2}\,dx=\sqrt{\pi}$$ which holds true since $$\int^\infty_{-\infty} e^{-x^2}\,dx=\sqrt{\pi}$$ 2- We need to prove the case $P(k)\to P(k+1)$ ...


1

Maybe I am being completely stupid, but why not simply write $$\Gamma(k + \tfrac{3}{2}) = \Gamma((k+1) + \tfrac{1}{2}) = \frac{\sqrt{\pi}(2(k+1))!}{4^{k+1}(k+1)!}, \quad k > -\frac{3}{2}?$$


0

Assume $n=2k+1$ is odd, so the integral is, by taking $t\to t/2$ $$\int_0^{\pi}\frac{\sin((2k+1)t)}{\sin t}dt= \int_0^{\pi/2}\frac{\sin(k+\frac 1 2)t}{2\sin \frac t 2}dt$$ The integrand is half of the Dirichlet kernel, which equals thus $$\frac 1 2 +\sum_{j=1}^k \cos jt$$ and makes the evaluation easy.


0

Use integration by parts, taking $\displaystyle \int_{0}^{x} (\arctan t)^2 dt$ as the first function and $ \dfrac{1}{x(x^2+1)} $ as the second function to get: $ \displaystyle \int \dfrac{dx}{x(x^2+1)} = \dfrac{1}{2} \ln\bigg(\dfrac{x^2}{x^2+1}\bigg)$ Hence, $I = \bigg|\dfrac{1}{2} \ln \bigg(\dfrac{x^2}{x^2+1}\bigg) \displaystyle \int_{0}^{x} (\arctan ...


3

If we substitute $w = e^{iz}$, we obtain $$\int_0^{2\pi} \frac{\cos z}{2+\cos z}\,dz = \int_{\lvert w\rvert = 1} \frac{\frac{1}{2}(w+w^{-1})}{2+\frac{1}{2}(w+w^{-1})}\, \frac{dw}{iw} = -i\int_{\lvert w\rvert = 1} \frac{w^2+1}{w(w^2+4w+1)}\,dw,$$ which can be evaluated directly with the residue theorem. We get something that can be evaluated with a little ...


1

First integrate with respect to $x$ i.e $$\int_0^1 \int_0^1 \int_0^1 x^{y^z}\,dx\,dy\,dz=\int_0^1 \int_0^1 \frac{1}{1+y^z}\,dy\,dz$$ Since $0<y<1$ and $0<z<1$, $$\int_0^1 \int_0^1 \frac{1}{1+y^z}\,dy\,dz=\sum_{k=0}^{\infty} (-1)^k \int_0^1 \int_0^1 y^{kz}\,dy\,dz=\sum_{k=0}^{\infty} \frac{(-1)^k\ln(1+k)}{k}$$ $$\begin{aligned} \sum_{k=0}^{\infty} ...


2

$$\displaystyle K=\int{\frac{dx}{(x^n+1)\large\sqrt[n]{x^n+1}}}$$ Let $$x^{n}=t \Rightarrow x=\large\sqrt[n]{t}\Rightarrow \normalsize dx=\frac{dt}{n\large\sqrt[n]{t^{n-1}}}$$ Hence, $$\displaystyle K=\int\frac{dt}{nt\large\sqrt[n]{\frac{t+1}{t}}}$$ Let $$\displaystyle \large \sqrt[n]{\frac{t+1}{t}}=\normalsize u\Rightarrow t=\frac{1}{u^{n}-1}\Rightarrow ...


2

Let $\displaystyle v=v(x)=\left(\frac{x^n+1}{x^n}\right)^{\Large \frac{1}{n}}$. Then we have: $\displaystyle v(0)=+\infty \ , \ v(1)=2^{\large \frac{1}{n}} \ , \ \frac{1}{x^n+1} = 1 - v^{-n} \ , \ x=\frac{1}{\left(v^n-1\right)^{\large \frac{1}{n}}}$ and $$\displaystyle dx = - \frac{v^{n-1}}{\left(v^n-1\right)^{1+\large \frac{1}{n}}} dv$$ Thus, ...


7

Your integral can be rewritten as \begin{align} \int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x(1-x)(1+x)}{\rm d}x =\int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x}{\rm d}x+\int^1_0\frac{x{\rm artanh}\ {x}\ln{x}}{1-x^2}{\rm d}x \end{align} The first integral is \begin{align} \int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x}{\rm d}x ...


1

Let $u = x^n + 1$, so that $dx = du/(nx^{n-1})$. $$\int_0^1{\frac{dx}{(x^n+1)\sqrt[n]{x^n+1}}} = \int_1^2{\frac{du}{nx^{n-1}(u\sqrt[n]{u})}} = \frac{1}{n}\int_1^2{\frac{du}{u\sqrt[n]{u}\sqrt[n]{(u - 1)^{n-1}}}}$$ $$= \frac{1}{n}\int_1^2{\frac{du}{u^{(n+1)/n}(u - 1)^{(n - 1)/n}}}$$ I believe this last integral can be resolved using partial fractions. I'll ...


3

Actually, we do need to worry about the pole at $x=1$ if we intend to use contour integration, for reasons that are a bit subtle. I will demonstrate below. The standard way to treat integrals of rational functions times logs over $[0,\infty)$ in complex analysis is to consider a keyhole contour, and an integral over that contour of the next higher power of ...


0

The easiest and certainly most general way is to compute $$PV\int_0^{\infty} dx \frac{x^a}{1-x^b},$$ and then take the derivative w.r.t. $a$ of the result. To do the first integral, use a circular sector ('pizza slice contour').


1

Here the measure space $X$ is the real line endowed with the Lebesgue measure. The set $[0,+\infty)$ is Borel measurable as the complement of the open set $(-\infty,0)$. a continuous function is indeed Borel-measurable.


2

As I said in comments, it seems that there is no explicit antiderivative. Even numerical methods could be problematic because the integrand is undefined at the bounds. One possible way is to perform a Taylor expansion around $x=0$. We so can get $$\frac{\tanh ^{-1}(x)}{(1-x) x (x+1)}=1+\frac{4 x^2}{3}+\frac{23 x^4}{15}+\frac{176 x^6}{105}+\frac{563 ...


1

METHOD Consider whether the Difference Function $\,f_3=f_1-f_2$ integrates to zero over the interval $0,2\pi$. If it does then $D_1 = D_2$. SUMMARY The difference function does integrate to zero over the given interval. Therefore the two definite integrals $D_1$ and $D_2$ are equal. DETAIL Let us try to prove the hypothesis that the two definite ...


0

Let $x = a\sin \theta$. Note that, $dx = a\cos \theta d\theta$. $$ \int_{0}^{a} x^2\sqrt{a^2 - x^2}dx = \int_{0}^{\pi/2} (a\sin \theta)^2\sqrt{a^2 - a^2\sin^2 \theta}\ a \cos \theta d\theta $$ $$ = \dfrac{a^4}{4}\int_{0}^{\pi/2} 4\sin^2\theta \cos^2 \theta d\theta = \dfrac{a^4}{4}\int_{0}^{\pi/2} \sin^2(2\theta) d \theta = \dfrac{a^4}{8}\int_{0}^{\pi/2}[1 - ...


1

Put $x=ay=a\sin \varphi$ Then $ \int_{0}^{a}dxx^{2}\sqrt{a^{2}-x^{2}}=a^{4}\int_{0}^{1}dyy^{2}\sqrt{1-y^{2}}% =a^{4}\int_{0}^{\pi /2}d\varphi \cos \varphi \sin ^{2}\varphi \cos \varphi $ I leave the rest to you.


2

First, use the change of variable $x=\frac{1}{u} => dx = -\frac{1}{u^2}du $ $I = \int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x = 2*\int_{ 0 }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x = 2*\int_{ 0 }^{ + \infty } {\frac{u^2} {1 + {u^4}}} \;{\mathrm{d}}u $ => $2*I = 2*\int_{ 0 }^{ + \infty } {\frac{1+x^2} {1 + {x^4}}} ...


0

Consider contour, which is semicircle (origin at $0$ and radius $R$), more precisely: $$\gamma=\{Re^{it}:t \in [0,\pi] \} \cup [-R,R]$$ According residue theorem: $$\oint_{\gamma}\frac{1}{x^4+1}dz=2\pi i (\text{Res}_{b}\frac{1}{x^4+1}+\text{Res}_{a}\frac{1}{x^4+1})$$ Where $a=e^{\pi i/4}$ and $b=e^{2\pi i /4}$ But on the other hand: ...


5

Since the integrand is an even function, we can rewrite it as \begin{equation} \int_{-\infty}^\infty\dfrac{1}{1+x^4}\ dx=2\int_{0}^\infty\dfrac{1}{1+x^4}\ dx \end{equation} In general, we have (click the formula below for the proof) \begin{equation} \int_0^\infty\dfrac{1}{1+x^n}\ dx=\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)},\qquad\mbox{for}\qquad n>0 ...


1

If I did this problem, I would first prove that $\int_{-\infty}^{\infty}\frac{1}{1+x^4}dx$ converges, then we have $$\int_{-\infty}^{\infty}\frac{1}{1+x^4}dx=\lim_{R\rightarrow\infty}\int_{-R}^{R}\frac{1}{1+x^4}dx=\lim_{R\rightarrow\infty}\left[\int_{C(R)^+}\frac{1}{1+z^4}dz-\int_{L(R)^+}\frac{1}{1+z^4}dz\right]$$ Where $L(R)$ is the upper part of the ...


1

use that $1+x^4=- \left( x\sqrt {2}-{x}^{2}-1 \right) \left( x\sqrt {2}+{x}^{2}+1 \right) $ and make a partial fraction decomposition


2

Your bounds are correct. Your integral in spherical coordinates becomes... $$ = \int_0^{2\pi} \int_0^{\pi/2} \int_0^{\cos(\phi)} \rho \cdot \rho^2 \sin(\phi) \,d\rho \,d\phi \,d\theta $$ Since nothing involves $\theta$, that integral can be factored out. $$ = \int_0^{2\pi}\,d\theta \int_0^{\pi/2} \int_0^{\cos(\phi)} \rho^3 \sin(\phi) \,d\rho \,d\phi = ...


4

Using substitution and integration by parts, \begin{align}&\int_0^1\left(\frac{\tan^{-1}x}{x}\right)^n\,dx=\int_0^{\pi/4}\frac{x^n}{\tan^n x}\sec^2 x\,dx\\&=\left[\frac{x^n}{\tan^n x}\tan x\right]_0^{\pi/4}-\int_0^{\pi/4}\frac{nx^{n-1}}{\tan^{n-1}x}-\frac{nx^n}{\tan^n x}\sec^2x\,dx\end{align} So we have ...


1

It is not the case that: $$ \int |g(x)| \, dx = \left|\int g(x) \, dx\right| $$ Instead, to get rid of the absolute values, we use the fact that: $$ |x - 1| = \begin{cases} x - 1 &\text{if } x \geq 1 \\ 1 - x &\text{if } x < 1 \end{cases} $$ We thus split the integral into two separate ones: $$ \int_{-5}^1 [(\tfrac{-1}{5}x + 7) - (2 + (1 - x))] \, ...


0

Well you could integrate $(f-|f|)/2$ (the negative part of the function), but it depends on $f$ if this leads to anything nice.


0

$f$ is continuous in $[-1, 1]$. Calculate $\int_0^1 f(2x -1) dx$, given that $\int_{-1}^1 f(u) du = 5$ $u = 2x - 1 \Rightarrow x = \frac{u+1}{2}$ $du = 2dx$ This is where I was confused: $$\int_{u = 2.0 - 1}^{u = 2.1 - 1} f(u) \frac{du}{2}$$ So: $\int_{-1}^1 f(u) \frac{du}{2} $ $\frac{1}{2}\int_{-1}^1 f(u) du = \frac{5}{2}$ I guess that if I face ...


5

Here is a simple and a nice way to evaluate the first and the second integral. Evaluation of $1^{\mbox{st}}$ Integral : Making substitution $x=\tan\theta\,$ followed by integration by parts, we get \begin{align} \int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx&=\color{red}{\int_0^{\Large\frac{\pi}{4}}\frac{\theta^2}{\sin^2\theta}\,d\theta}\\ ...


0

Counterexample: $n=1$, $m=-1$, $$\int_{-\pi}^{\pi}\,\cos(mx)\cos(nx)\,dx = \pi$$


7

For the first one, \begin{align} \int^1_0\frac{\arctan^2{x}}{x^2}{\rm d}x =&\color{#BF00FF}{\int^\frac{\pi}{4}_0x^2\csc^2{x}\ {\rm d}x}\\ =&-x^2\cot{x}\Bigg{|}^\frac{\pi}{4}_0+2\int^\frac{\pi}{4}_0x\cot{x}\ {\rm d}x\\ =&-\frac{\pi^2}{16}+4\sum^\infty_{n=1}\int^\frac{\pi}{4}_0x\sin(2nx)\ {\rm d}x\\ ...



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