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17

$$I:=\int_{0}^{\infty}\frac{\ln{(x)}}{\sqrt{x}\,\sqrt{x+1}\,\sqrt{2x+1}}\mathrm{d}x.$$ After first multiplying and dividing the integrand by 2, substitute $x=\frac{t}{2}$: ...


11

Recall: $$x^2-xy+y^2=(x-\frac{1}{2}y)^2+\frac{3}{4}y^2$$ With that you get: $$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}(x^2-xy+y^2)}dxdy=\\\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left((x-\frac{1}{2}y)^2+\frac{3}{4}y^2\right)}dxdy=\\ \int_{-\infty}^\infty\int_{-\infty}^\infty ...


10

We will prove that $$I=-\frac{\pi^4}{2880}.$$ Indeed, let $$ J=\int_0^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx $$ It is easy to see that $$\eqalign{J&=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+ \int_{\pi/4}^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx\cr &=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+ \int_{0}^{\pi/4}\log^2(\cos ...


9

Here is an approach without using contour integration (Cauchy theorem). I've found the following general result. Theorem 1. Let $a$ be any real number. Then $$ \begin{align} \displaystyle \int_{0}^{\pi} \frac{x}{x^2+\ln^2(2 e^{a} \sin x)}\mathrm{d}x & = \, \frac{2 \pi^2}{\pi^2+4a^2},\tag1\\\\ \int_{0}^{\pi} \frac{\ln(2 e^{a} \sin ...


8

I also tried expanding it with a series, but it didn't help. It should have helped. If we write $$\frac{x^{-a}-x^a}{1-x} = (x^{-a}-x^a)\sum_{n=0}^\infty x^n,$$ by the monotone convergence theorem, we have $$\begin{align} \int_0^1 \frac{x^{-a}-x^a}{1-x}\,dx &= \sum_{n=0}^\infty \int_0^1 (x^{-a}-x^a)x^n\,dx\\ &= \sum_{n=0}^\infty ...


8

With the substitution $x^2=t$, $$I=\frac{1}{2}\int_0^{\infty} \frac{\sin^3(\pi t)\cos(4t)}{t^3}\,dt$$ Since, $$\sin(3x)=3\sin x-4\sin^3 x \Rightarrow \sin^3x=\frac{3\sin x-\sin(3x)}{4}$$ $$\Rightarrow \sin^3(\pi t)=\frac{3\sin(\pi t)-\sin(3\pi t)}{4}$$ i.e $$\begin{aligned} I &=\frac{1}{8}\int_0^{\infty} \frac{(3\sin(\pi t)-\sin(3\pi ...


8

Use polar coordinates, We know that $$r^2 = x^2 + y^2$$ So our double integral becomes $$\int_{0}^{2\pi} \int_0^{1}r^2\cdot rdrd\theta$$ Now solve. EDIT I see that your computation is correct, I am simply offering another alternative and more easier way to solve this double integral.


7

Hint : Try the substitution $$t=\cfrac{T}{u^2+1}$$ The first integral has the shape of the gaussian. The second one leads you to $$I(\beta) = \alpha \int_{\mathbb{R}^+} \cfrac{1}{u^2+1} \exp(-\beta(u^2+1) ) \,du$$ Considering $$\begin{cases} I'(\beta)=\alpha \int_{\mathbb{R}^+} \exp(-\beta(u^2+1) ) \,du=\alpha\exp(-\beta)\cfrac{\sqrt{\pi}}{2\sqrt{\beta}} ...


7

Note that $$\int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta= \int_{- \pi /2}^{\pi / 2} \sin^2 \theta d \theta \ $$ and that $$\pi = \int_{- \pi /2}^{\pi / 2} 1 d \theta = \int_{- \pi /2}^{\pi / 2} (\cos^2 \theta +\sin^2 \theta) d \theta = 2 \int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta$$ hence $$\int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta= ...


6

I am afraid that what you did is wrong : you are not integrating a polynomial expression. Just for your curiosity,$$\frac{d}{dt}\Big(\frac{e^{t+t^2}}{t^2/2+t^3/3}\Big)=\frac{6 e^{t+t^2} (t+2) \left(4 t^2-3\right)}{t^3 (2 t+3)^2} \neq e^{t+t^2}$$ To compute $$I=\int e^{t}.e^{t^2} dt=\int e^{t^2+t} dt$$ first complete the square for the exponent and perform a ...


6

Since no answers have been posted, I'll expand on my comment above. There is a general formula that states $$\sum_{n=1}^{\infty}\frac{H_{n}^{(r)}}{n^{q}}=\zeta(r)\zeta(q)-\frac{(-1)^{r-1}}{(r-1)!}\int_{0}^{1}\frac{\text{Li}_{q}(x) \log^{r-1}(x) }{1-x}dx $$ where $$ H_{n}^{(r)} = \sum_{k=1}^{n} \frac{1}{k^{r}} .$$ A proof can be found here. Making the ...


6

If you use polar coordinates you will get the following $$ \int_{0}^{2\pi} \int_{0}^{1} r^2 r dr \ d\theta $$


5

We have $\lim_{y\to-\infty}\phi(y)=0$ and $\lim_{y\to\infty}\phi(y)=1$. Moreover, the function $\phi$ is continuous. Thus, by the Intermediate Value Theorem, the function $\phi$ is surjective.


5

We have: $$ J=\int_{0}^{+\infty}\frac{x^p}{1+x^2}\arctan(x)dx,$$ but since $\arctan x=\Im\log(1+xi)=-\Im\log(1-xi)$, $$ J=\frac{1}{2}\Im\int_{0}^{+\infty}\frac{x^p}{1+xi}\log(1+xi)\,dx-\frac{1}{2}\Im\int_{0}^{+\infty}\frac{x^p}{1-xi}\log(1-xi)\,dx$$ hence: $$ ...


5

The main idea is that, near the largest point of the integrand (which occurs at $x=0$), we have $$ \frac{1}{\log x} + \frac{1}{1-x} \approx \frac{1}{\log x} + 1. $$ So we split the integral up as $$ \int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx = \left[\int_0^{1/e} + \int_{1/e}^1 \right] \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx. ...


5

Use the substitution: $x = \tan\theta$. The integral is then equal to: $$I= \int_{0}^{\pi/4} \ln(1+\tan\theta) \ d\theta (*)$$ Also,we know the property: $$\int_{0}^{b} f(x) \ dx = \int_{0}^{b} f(b-x) \ dx$$ so we have $$I = \int_{0}^{\pi/4} \ln\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \ d\theta = \int_{0}^{\pi/4} ...


4

The general formula for the average value of a function $f(x)$ over $a \le x \le b$ is $$f_{\text{avg}} = \dfrac{1}{b-a}\displaystyle\int_a^bf(x)\,dx$$ Here, you want the avgerage value of the function $f(x) = \tan 2x$ over $0 \le x \le \dfrac{\pi}{8}$. Can you use the above formula to get the answer? EDIT: Since you are having trouble integrating $\tan ...


4

If $f$ is decreasing on $[r,r+1]$ then $f(r) > f(t)$ for $r < t \le r+1$, which yields: $f(r) = \displaystyle\int_{r}^{r+1}f(r)\,dt > \int_{r}^{r+1}f(t)\,dt$. So, the inequality is false for any decreasing function, such as $f(t) = \dfrac{1}{(1+t)^2}$


4

I highly doubt that one can evaluate that integral by hand, however the integral does have a closed form found by Mathematica to be: $$\frac{2 \left(4 (-1)^{5/6} \, _4F_3\left(1,1,1,\frac{7}{6};2,2,2;\frac{1}{9}\right)+8 (-1)^{5/6} \log (3) \, _3F_2\left(1,1,\frac{7}{6};2,2;\frac{1}{9}\right)+432 (-1)^{5/6} \log ^2(2)-189 (-1)^{5/6} \log ^2(3)+648 (-1)^{5/6} ...


4

Too long for a comment and this is not the answer too, but following @David H's comment. Perhaps your brother means $$\int_{\color{red}{\large\frac12}}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\ dx.\tag1$$ If so, then setting $u=2x-1$ to $(1)$ yields $$ \frac1{\sqrt[\large 3]{4}}\int_{0}^1\frac{\ln u}{\sqrt[\large 6]{(1+u)(1-u)u^4}}\ ...


4

Integrating by parts 3 times, $$ \begin{align} \int_{0}^{\pi /2} \frac{x^{4}}{\sin^{4} x} \ dx &= - \frac{x^{4}}{3} \cot(x) \left(\csc^{2} (x) +2 \right) \Bigg|^{\pi/2}_{0} + \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot (x) \left(\csc^{2} (x) +2 \right) \ dx \\ &= \frac{4}{3} \int_{0}^{\pi /2} x^{3} \cot (x) \left(\csc^{2} (x) +2 \right) \ dx \\ &= ...


4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...


4

Consider $$ \mathcal{I}(y,t)=\int_{-\infty}^{\infty}\frac{\cos xt}{x^2+y^2}\ dx=\frac{\pi e^{-yt}}{y}\quad;\quad\text{for}\ t>0.\tag1 $$ Differentiating $(1)$ with respect $t$ and $y$ yields \begin{align} \frac{\partial^2\mathcal{I}}{\partial y\partial t}=\int_{-\infty}^{\infty}\frac{2xy\sin xt}{(x^2+y^2)^2}\ dx&=\pi te^{-yt}\\ ...


4

From the hint, $\cos^2 x= \frac{1}{2}(\cos 2x+1)$ so our integral becomes $$\int_{-\pi/2}^{\pi/2} \frac{1}{2}(\cos 2x+1)dx= \left. \left (\frac{1}{4}\sin 2x+\frac{1}{2}x \right ) \right |_{-\pi/2}^{\pi/2}=\pi/2$$


4

We can try the harmonic analysis path. Since: $$\log(2\cos x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n}\cos(2nx),$$ $$\log(2\sin x)=-\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{n},\tag{1}$$ we have, as an example: $$\int_{0}^{\pi/2}\log^3(2\sin x)dx=-\frac{3\pi}{4}\zeta(3)$$ since $$\int_{0}^{\pi/2}\cos(2n_1 x)\cos(2n_2 x)\cos(2n_3 x)\,dx = ...


4

Proposition. $$ \int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:{\mathrm{d}}x = \frac{45}{512} \zeta(7)-\frac{3\pi^2}{16} \zeta(5)-\frac{5\pi^4}{64} \zeta(3)-\frac{9}{4}\zeta(\bar{5},1,1) $$ where $\zeta(\bar{p},1,1)$ is the colored MZV (Multi Zeta Values) function of depth 3 and weight $p+2$ given by $$ \zeta(\bar{p},1,1) : = \sum_{n=1}^{\infty} ...


3

You can use a comparison to show that a smaller integral diverges: $$ \left.\int\limits_\epsilon^\infty \frac{1}{x}dx = \ln(x)\right|_\epsilon^\infty \longrightarrow \infty $$ Since $\frac{1}{x} \leq \frac{\ln(x)}{x}$ for all $x \geq e$, we have: $$ \int\limits_e^\infty \frac{1}{x}dx \leq \int\limits_e^\infty \frac{\ln(x)}{x}dx $$ Since the smaller ...


3

Let $l=r\tan{u}$, then $dl=r\sec^2{u} \ du$. The integral becomes $$\frac{1}{r^2}\int^{\pi/2}_0\frac{\sec^2{u}}{\sec^3{u}}du=\frac{1}{r^2}$$


3

Using $k=\frac{1}{2}$ in the second rule, we get $$\begin{align} \int\limits_{-2}^{2} (x-3) \sqrt{4-x^2} \ dx &=2\int\limits_{-1}^{1} (2x-3) \sqrt{4-4x^2} \ dx \\&=2\int\limits_{-1}^{1} 2(2x-3) \sqrt{1-x^2} \ dx \\&=4\int\limits_{-1}^{1} (2x-3) \sqrt{1-x^2} \ dx \\&=8\int\limits_{-1}^{1}x\sqrt{1-x^2} \ dx-12\int\limits_{-1}^{1}\sqrt{1-x^2}\ ...


3

$$ \int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta + \int_{-\pi/2}^{\pi/2} \sin^2\theta\,d\theta = \int_{-\pi/2}^{\pi/2} 1\,d\theta=\pi. $$ If you can show the two integrals are equal, then they each have to be $\pi/2$. But they have to be equal since the graph of $\sin^2$ has the same size and shape as that of $\cos^2$ and the interval from $-\pi/2$ to $\pi/2$ ...



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