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20

For any $\alpha > 0$ and $n \in \mathbb{Z}_{+}$. Let $I_n(\alpha)$ be the $n$-dimensional integral $$I_n(\alpha) = \int_{[0,1]^n} \left\{1\left/\prod_{i=1}^n x_i\right.\right\}^\alpha \prod_{i=1}^n dx_i$$ It is clear all these $I_n(\alpha) \in (0,1)$. As a result, the corresponding generating function $$I(\alpha,z) = \sum_{k=0}^\infty ...


12

This integral is not equal to zero. We may obtain the following closed form. $$ \begin{align} \int_0^1 \left(\left\{\frac{1}{x}\right\}-\frac{1}{2}\right)\frac{\log(x)}{x} \mathrm{d}x & = \dfrac{\ln^2(2\pi)}{4}-\dfrac{\gamma^2}{4}+\dfrac{\pi^2}{48}-\dfrac{\gamma_1}{2}-1\tag1 \\\\ \end{align} $$ where $\left\{x\right\}$ denotes the fractional ...


9

Here is my approach. Theorem 1. Let $s$ be a complex number such that $-1<\Re{s}<1$, $s\neq 0$. Then $$ \int_{0}^{1} x^{s}\left\{1/x\right\}^{2}\:\mathrm{d}x = -\frac{2\zeta(s)}{s(1+s)}-\frac{\zeta(1+s)}{1+s}-\frac{1}{1-s} \tag1 $$ where $\left\{x\right\}=x-\lfloor x\rfloor$ denotes the fractional part of $x$ and where $\zeta$ denotes the ...


9

The integral as you've written isn't well-defined - what you've written is: $\qquad$Integrate $\frac1x$ along the line running from $-1$ to $+1$ However, since $\frac1x$ isn't defined at 0, this doesn't make sense. As such, the integral as you've written it is slightly ambiguous. The answer to whether or not the integral converges will depend on how you ...


7

WA is probably summing the well-defined integrals $$\int_0^1\log x\,\mathrm dx=\left.x\log x-x\right|_0^1=-1$$ and $$\int_{-1}^0\log x\,\mathrm dx,$$ using the convention that, when $x$ is real and negative, $\log x=\mathrm i\pi+\log|x|$, hence $$\int_{-1}^0\log x\,\mathrm dx=\int_0^1(\mathrm i\pi+\log u)\,\mathrm du=\mathrm i\pi-1,$$ by the first ...


6

$$I=\frac\pi{\sqrt[4]2}+\sqrt{20-14\sqrt2}\ K\!\left(2\sqrt2-3\right)+\sqrt{2+\sqrt2}\ E\!\left(2\sqrt2-3\right)\\-2\sqrt{2-\sqrt2}\ \Pi\left(\sqrt2-1,\,2\sqrt2-3\right)$$ where $K(m), E(m), \Pi(n,m)$ are the complete elliptic integrals of the first, second and third kind: $$K(m)={\large\int}_0^{\pi/2}\frac{d\theta}{\sqrt{1-m\sin^2\theta}}$$ ...


5

With $t=\cos(x)$, we see it is an elliptic integral, $$ I = \int_{1/\sqrt{2}}^1 \frac{dt}{t^2\sqrt{(1-t^2)(t-1/\sqrt{2})}} $$ added Maple gets: if $0<a<1$, then $$ \int _{a}^{1}\!{\frac {dt}{{t}^{2}\sqrt { \left( 1-t^2 \right) \left( t-a \right) }}}{} =-{\frac {\sqrt {2}{\bf K} \left( 1 /2\,\sqrt {-2\,a+2} \right) }{a}} +{\frac {\sqrt {2} ...


5

First note that: $$I=\int_{0}^{1}\frac{\mathrm{d}x}{(1+x^2)\sqrt{1+x^2}}=\int_{0}^{1}\frac{\mathrm{d}x}{(1+x^2)^{3/2}}.$$ If you then had the presence of mind to substitute $u=\frac{x}{\sqrt{1+x^2}}$, you would notice that $$\mathrm{d}u=\frac{\mathrm{d}x}{(1+x^2)^{3/2}},$$ and thus, $$I=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}u=\frac{1}{\sqrt{2}}.$$


5

You can compute the integral directly using complex coordinates. Let $z = x + iy$ and $\bar{z} = x-iy$ and notice $$dx \wedge dy = \frac{d\bar{z} \wedge dz}{2i}$$ The integral $I_k$ over the disk $B_k$ becomes $$\int_{B_k} \frac{4}{|1 + (x+iy)|^4} dx \wedge dy = -2i \int_{B_k} \frac{1}{(1+z)^2(1+\bar{z})^2} d\bar{z} \wedge dz\\ = 2i \int_{B_k} d \left( ...


5

The main issue here is that, despite the odd-ness of the integrand function, $\frac{1}{x}$ is not a Riemann integrable function over $(0,1)$, hence the value of such integral, if existing, would be something like $\infty-\infty$. Also notice that: $$\int_{(-1,-a/n)\cup(b/n,1)}\frac{dx}{x}=-\log\frac{n}{a}+\log\frac{n}{b}=\log\frac{a}{b}.$$ By choosing ...


5

Let $A=a+b$ and $B=a-b$. Convert this into a complex integral and integrate over a semicircle in the uhp. The poles that lie in the contour are $\displaystyle z={\rm{i}}$ and $z={\rm{i}}\sqrt{B/A}$. As the radius $\to \infty$ the integral over the arc vanishes, hence for $\color\red{a>b>0}$, \begin{align} I ...


5

I am going to go ahead and post my method. It is similar to xpauls except I used digamma, which is related to the harmonic series anyway. Break integral up: $$\int_{0}^{1}\frac{\log^{2}(x)\log(1+x)}{(1-x)(x^{2}+1)}dx+\int_{1}^{\infty}\frac{\log^{2}(x)\log(1+x)}{(1-x)(x^{2}+1)}dx$$ In the right integral, make the sub $x=1/t$. This gives: ...


4

The integral is evaluated in terms of the Gamma function as follows $${\frac {\sqrt {\pi }\Gamma \left( -1/2+\alpha \right) }{\Gamma \left( \alpha \right) }} $$ Then $\alpha >1/2$ is domain with convergence. When $\alpha =1/2$ the integral does not converge. When $-1/2 + \alpha = -n$ where $n$ is a positive integer, the integral does not converge. ...


3

Outline: Use Comparison. Break up into two parts at $0$. By symmetry the two integrals both exist or both fail to exist, and if they exist they have the same value. Since our function is nicely behaved on the interval $[-1,1]$, we only need to worry about $\int_1^\infty \frac{1}{(1+x^2)^\alpha}\,dx$. For $\alpha \gt 1/2$, we have convergence, by ...


3

Hint: what about differentiating both sides and solving for $x(t)$ (provided $x(t)$ is differentiable - thanks to @LiuGang for pointing out)? You will need to recall that: $$ \color{blue}{\frac{\mathrm{d} }{\mathrm{d} t} \int^t_{a} f(s) \, \mathrm{d}s =f(t), \quad a \in \mathbb{R}}$$ and, of course, that: $$\color{blue}{ x'(t) + c \, x = 0 ...


3

Here are a couple of excerpts that might be of use to you or others in tracking down more information about this topic: Joseph Fels Ritt (1893-1951), Integration in Finite Terms. Liouville's Theory of Elementary Methods, Columbia University Press, 1948, ix + 100 pages. from p. 60: We should like to present a class of problems in which numbers are ...


3

Instead of writing a formula in summation notation of $\ln(1+x)$, I personally find it easier to expand the Taylor series expansion of $\ln (1+x)$, like this: $$\ln (1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\frac{x^5}5-\cdots,$$ where the expansion is sufficiently long enough to help you integrate and find your answer precisely to at least 2 decimal ...


3

Note that the terms in your series fall very rapidly to zero. Since it is an alternating sum the error when summing up to and including term $N$ is bounded by the absolute value of term $N+1$. You will not need many terms to get two digits precision.


3

If you take the integral from $0$ to $\pi$ and then multiply by $70$, you've got it, since it's periodic with period $\pi$. On the interval from $0$ to $\pi$ you can drop the absolute value since the function is nonnegative there. And $$ \int_0^{\pi/2} (\cos^2 x) \Big(\sin x\,dx\Big) = \int_1^0 u^2 \Big(-du\Big). $$ And finally, ...


3

Let $z=e^{it}$, then $dz=ie^{it}\ dt$ or $dt=\dfrac{dz}{iz}$, and $\cos t=\dfrac{e^{it}+e^{-it}}{2}=\dfrac{z+z^{-1}}{2}$. \begin{align} \int_0^{2\pi}\frac{1}{4+\cos t}dt&=\oint_C\frac{1}{4+\frac{z+z^{-1}}{2}}\cdot\frac{dz}{iz}\\ &=\frac2i\oint_C\frac{1}{z^2+8z+1}dz, \end{align} where $C$ is the circle of unit radius with its center at the origin. The ...


2

However, this integral doesn't need more attention; you could also use the substitution to solve the indefinite associated integral: $$1+x^2=t^2x^2$$ And soyou'll find the integral as $$\int(-1/t^2)dt$$


2

Note that your integral can be rewritten as follows: $$\color{blue}{ I = \int^{2\pi}_0 \frac{2}{8+ e^{it}+ e^{-it}} \, \mathrm{d}t = \int^{2\pi}_0 \frac{2 e^{it}}{e^{i 2 t} + 8 e^{it} + 1} \, \mathrm{d}t}$$ Define now $z \equiv e^{it}$, $\mathrm{d}z = ie^{it} \, \mathrm{d}t = i z \, \mathrm{d}t $, so: $$ \color{blue}{I = \frac{2}{i} \int_\gamma ...


2

The integrand function is constant on a sequence of intervals, in particular: $$I=\int_{0}^{1}\frac{dx}{\left\lfloor 1-\log_2 x\right\rfloor}=\sum_{n=1}^{+\infty}\frac{2^{1-n}-2^{-n}}{n}=\sum_{n=1}^{\infty}\frac{2^{-n}}{n}=\color{red}{\log 2.}$$


2

As already answered by Troy Woo, partial fraction decomposition gives $$ \dfrac{1-t^2}{(1+t^2)\Big((a+b)t^2+(a-b)\Big)}=\frac{a}{b\Big( (a+b) t^2+(a-b)\Big)}-\frac{1}{b \left(t^2+1\right)}$$ which lead to simple integrals. As a result $$\int \dfrac{1-t^2}{(1+t^2)\Big((a+b)t^2+(a-b)\Big)} dt= \frac{a \tan ^{-1}\left(\frac{t \sqrt{a+b}}{\sqrt{a-b}}\right)}{b ...


2

Apply partial fractional decomposition, and then you will easily solve it with basic techniques.


2

We assume that $R$ is positive. Let $y=R\tan\theta$ or equivalently $\theta=\arctan(y/R)$. As $y$ ranges from $0$ to $\infty$, the number $\theta$ ranges from $0$ to $\pi/2$. Note that $\sqrt{R^2+y^2}\left(R^2+y^2\right)= R^3 \sec^3\theta$ and $dy=R\sec^2\theta\,d\theta$. We leave the rest to you.


2

Using $y=Rx$, then $x=\tan(\theta)$ we get $$ \begin{align} \int_0^\infty\frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}\mathrm{d}y &=\frac1R\int_0^\infty\frac{\mathrm{d}x}{\sqrt{1+x^2}\left(1+x^2\right)}\\ &=\frac1R\int_0^{\pi/2}\cos(\theta)\,\mathrm{d}\theta\\ &=\frac1R \end{align} $$ It seems that you have miscomputed the primitive. Computing ...


2

$$I=\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G,$$ where $A$ is the Glaisher–Kinkelin constant and $G$ is the Catalan constant.


1

By simetry, then $$\int_0^{70 \pi} \left|\cos^{2}\!\left(x\right)\sin\!\left(x\right)\right| dx=70\cdot\int_0^{ \pi} \cos^{2}\!\left(x\right)\sin\!\left(x\right) dx$$


1

Related problems (I). Here is how you can evaluate the integral. Making the change of variables $u=x-1$ gives $$I = \int_{0}^{\infty}\frac{\cos(\tau u)}{(1+u)^{5/3}}du. $$ Using the Taylor series of $\cos(\tau u)$ we get $$ I = \sum_{k=0}^{\infty}\frac{(-1)^k \tau^{2k} }{(2k)!} \int_{0}^{\infty}\frac{u^{2k}}{(1+u)^{5/3}}du. $$ Making the change ...



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