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7

Our integral equals: $$\begin{eqnarray*} I=\sum_{n\geq 0}\iint_{[0,1]^2}x^n y^n(1-x)^n (1-y)^n\,dx\,dy &=& \sum_{n\geq 0}\left(\int_{0}^{1}x^n(1-x)^n\,dx\right)^2\\&=&\sum_{n\geq 0}\left(\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}\right)^2\\&=&\frac{1}{4}\sum_{n\geq 0}\frac{1}{(n+1)^2\binom{2n}{n}^2}\tag{1}\end{eqnarray*}$$ and: $$ \sum_{n\geq ...


6

Let $x=\sin^2(t)$ and $y=\sin^2(w)$, we obtain \begin{align} I & = \int_0^{\pi/2}\int_0^{\pi/2} \dfrac{4\sin(t)\cos(t)\sin(w)\cos(w)dtdw}{1-\sin^2(t)\cos^2(t)\sin^2(w)\cos^2(w)}\\ & = 4\sum_{k=0}^{\infty}\int_0^{\pi/2}\int_0^{\pi/2} \sin^{(2k+1)}(t) \cos^{(2k+1)}(t)\sin^{(2k+1)}(w) \cos^{(2k+1)}(w)dtdw\\ & = 4 \sum_{k=0}^{\infty} ...


6

An incomplete answer. Expand as a geometric series and use the beta function. \begin{align} \int^1_0\int^1_0\frac{{\rm d}x\ {\rm d}y}{1-xy(1-x)(1-y)} &=\sum^\infty_{n=0}\left(\int^1_0x^n(1-x)^n\ {\rm d}x\right)^2\\ &=\sum^\infty_{n=0}\frac{\Gamma^4(n+1)}{\Gamma^2(2n+2)}\\ &=\sum^\infty_{n=0}\frac{1}{(2n+1)^2\binom{2n}{n}^2}\\ \end{align} Next, ...


6

$$y=(x^{n}+(1-x)^{n})^{\frac{1}{n}}=(1-x)\left(1+\left(\frac{x}{1-x}\right)^{n}\right)^{\frac{1}{n}}$$ Now, in $0\leq x \leq \frac{1}{2}$ $$0\leq \frac{x}{1-x} \leq 1$$ $$0\leq \left(\frac{x}{1-x}\right)^{n} \leq 1$$ $$1\leq \left(1+\left(\frac{x}{1-x}\right)^{n}\right)^{\frac{1}{n}} \leq 2^{\frac{1}{n}}$$ $$(1-x)\leq y \leq 2^{\frac{1}{n}}(1-x)$$ ...


5

The integral does not possess any elementary anti-derivative: see Liouville's theorem and the Risch algorithm for more information. It can be rewritten in terms of elliptic integrals or hypergeometric functions, by expanding the integrand into its own binomial series, and then reversing the order of summation and integration.


4

Set $x^2-1 = t$, we then have $2xdx = dt$, which gives us $$x^3 = \dfrac{x^2 (2xdx)}2 = \dfrac{(1+t)dt}2$$ Hence, the integral becomes $$\int \dfrac{x^3}{\sqrt{x^2-1}}dx = \int \dfrac{(1+t)dt}{2\sqrt{t}} = \sqrt{t} + \dfrac{t^{3/2}}3 + \text{ constant} = \sqrt{x^2-1} + \dfrac{(x^2-1)^{3/2}}3 + \text{ constant}$$


4

Note that, if $u^2=x^2-1$ this does not implies that $du=dx$. Indeed, you have to proceed as follows: $u^2=x^2-1$ gives $ u=\sqrt{x^2-1} $, and so $du =\frac{xdx}{\sqrt{x^2-1}}$. this implies that your integral become $$\int_{1}^{2} u^2+1 du = \frac{8}{3}+2-\frac{1}{3}-1 = \frac{10}{3}$$


4

You need to substitute every instance of $x$. You are then at $$ \int \frac{U+3}2 U^4 \frac{dU}2$$ and then use the formula $$ \int t^m dt = \frac {t^{m+1}}{m+1} $$ Can you take it from here?


4

First evaluate $$\int_{-2y}^{2y}x^2y^2dx$$ treating $y$ as some contant with a value between $0$ and $1$. You will get an expresion on $y$, which now you should integrate between $0$ and $1$, say $$\int_0^1 f(y)dy$$


4

$$ \iint x^2y^2\,dx\,dy=\int_0^1y^2\left[\int_{-2y}^{2y}x^2\,dx\right]\,dy=\int_0^1y^2\left[\frac{x^3}{3}\right]_{-2y}^{2y}\,dy=\int_0^1\frac{16}{3}y^5\,dy=\frac{16}{3\cdot6}y^6\Big|_0^1=\frac89. $$


4

This integral can be evaluates with calculus only: \begin{align*} \int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3} \, dx &= \frac{1}{a^{3/2}} \int_{0}^{\infty} \frac{x^2 + 1}{x^6 + 1} \, dx \tag{1} \\ &= \frac{1}{a^{3/2}} \int_{0}^{\infty} \frac{1}{(x - \frac{1}{x})^2 + 1} \, \frac{dx}{x^2} \\ &= \frac{1}{2a^{3/2}} \int_{0}^{\infty} ...


3

Thanks Achille Hui for the hint. Adding the answer for future reference of users. $$\int \limits^{\infty }_{0}\frac{dx}{a^{2}+(x-\frac{1}{x})^{2}} =\frac{\pi}{5050}$$ Hint: $$\int_0^\infty dx = \left(\int_0^1 + \int_1^\infty\right) dx = \int_1^\infty \left(\frac{dx}{x^2} + dx\right) = \int_0^\infty dy\quad\text{ with }\quad y = x - \frac{1}{x}$$ ...


3

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{a^2+x^2+y^2}dxdy=\int_{0}^{2\pi}\int_{0}^{\infty}\frac{1}{a^2+r^2}rdrd\theta$$ To appropriately transform coordinates from $x-y$ to $r-\theta$, where $x=r\cos \theta$ and $y=r\sin \theta$, we need to determine the determinant of the Jacobian matrix $J$. To that end, we have $$\begin{align} ...


3

For the region where $x$ goes from $x=0$ to $x=1$ $y$ goes from $4-2x$ to $4$. For the region where $x$ goes from $x=1$ to $x=4$ $y$ goes from $2\sqrt{x}$ to $4$. Thus, the area is given by $$\text{Area} =\int_0^1 \int_{4-2x}^4 dydx+\int_1^4 \int_{2\sqrt x}^4 dydx$$ Can you finish?


3

Simplify your boundary equations: $$y = 2 \sqrt{x}$$ $$y = 4$$ $$y = -2x +4$$ Sketch the area. You ought to try hand-sketching it to verify. Split into two double integrals. $$Area = \int_{x=0}^1\int_{y=lower curve}^{higher curve} dydx+ \int_{x=1}^4\int_{y=lower curve}^{higher curve}dydx$$. UPDATE/EDIT: You ought to have solved it by now. For ...


2

Here's the region: You can perform the integration in two parts, where the limits are determined by the values of $x$ where the appropriate two equations have the same value. I'll leave the details of that exercise to you but the result is ${11 \over 3}$. Since you asked, here's the Mathematica code: Plot[{4, 2 Sqrt[x], If[0 < x < 1, 4, Null], ...


2

If $x=e^u$, then $$ \int (\log x)^2 \, dx = \int u^2 e^u \, du. $$


2

We have: $x=e^u \implies dx = e^u\,du$ This gives you the integral $$\int u^2e^u\,du$$ Integration by parts, twice, should do the trick.


2

expand $(x-a)^n$ using the binomial theorem, and then integrate termwise using: $$ \int_0^\infty x^ne^{-x} dx = n! $$


2

Note that the given function $f$ is actually continuous, as the left- and right-hand limits of $f(x)$ at $x=1$ are equal to $1$. Thus, you can also say that $f(x) = x$ for $x$ in the closed interval $[-3, 1]$. This wouldn't actually be a problem even in the case of a jump discontinuity in your piecewise function. If you dive down into working with the ...


2

Using capital $U$ and lower-case $u$ interchangeably gets perceived by mathematicians the way others perceive a spelling error, and moreover in many contexts the corresponding capital and lower-case letters refer to two different things in the same problem. You've probably seen this in trigonometry where $a$ is the length of the side of a triangle and $A$ ...


2

In the same spirit as Mookid's answer, consider a more general case $$I=\int (A+Bx)(C+Dx)^n\,dx$$ Change variable $C+Dx=u$, $x=\frac{u-C}{D}$, $dx=\frac{du}{D}$ $$I=\int\big(A+\frac{B (u-C)}{D}\big)u^n \,du=\int\big(A-\frac{B C}{D}+\frac{B u}{D})u^n\,du$$ $$I=\big(A-\frac{B C}{D}\big)\int u^n \, du+\frac{B }{D}\int u^{n+1} \, du$$ $$I=\big(A-\frac{B ...


2

There is another step using integration by part $$\int \frac{x^3}{\sqrt{x^2-1}}dx = \int x^2\frac{x}{\sqrt{x^2-1}}dx$$ You want to integrate $x(x^2-1)^{-1/2}$ (and differentiate $x^2$), consider product rule. (Substitution is using the idea of product rule, i.e. you can let $u=x^2-1$) $$\int x(x^2-1)^{-1/2} = (x^2 - 1)^{1/2}dx$$ So the initial ...


2

I know what is happening now. Here is the general result: Let $f(z)$ be a rational function with real coefficients and no real poles which vanishes to order $2$ at $x=\infty$. Let $a_1 < b_1 < a_2 < b_2 < \cdots < a_{n-1} < b_{n-1} < a_n$ be real numbers and set $r(x) = c \frac{\prod(x-a_i)}{\prod(x-b_i)}$ for $c>0$. Then ...


2

$\int_0^\infty e^{-\frac{At^2}{t+1}}~dt$ $=\int_1^\infty e^{-\frac{A(t-1)^2}{t}}~d(t-1)$ $=\int_1^\infty e^{-\frac{A(t^2-2t+1)}{t}}~dt$ $=e^{2A}\int_1^\infty e^{-A\left(t+\frac{1}{t}\right)}~dt$ $=e^{2A}K_{-1}(A,A)$ (according to http://artax.karlin.mff.cuni.cz/r-help/library/DistributionUtils/html/incompleteBesselK.html)


2

There are several symmetries that you can take advantage of here. First of all, we can integrate $$\int_{-5}^5 \int_{-5}^5 15 \:\mathrm{d}x \:\mathrm{d}y = 15 \cdot (10\cdot 10) = 1500.$$ The $(10\cdot 10)$ represented the area of the square $[-5,5]\times [-5,5]$. As for what's left, $f(x,y) =|x+y|+|x-y|$ has certain symmetries. First of all $f(x,-y) = ...


2

For the second one, move to spherical coordinates, i.e., $x = r\sin(\theta)\cos(\phi)$, $y = r\sin(\theta)\sin(\phi)$ and $z = r\cos(\theta)$. We then have $$dxdydz = r^2\sin(\theta)drd\theta d\phi$$ Hence, the integral becomes \begin{align} \int_0^{2\pi}\int_0^{\pi} \int_0^1 \log(1/r)r^2\sin(\theta)drd\theta d\phi & = \left(\int_0^{2\pi}d\phi\right) ...


1

You are integrating over a $10 \times 10$ square centered on the origin. If you draw the diagonal from $(-5,-5)$ to $(5,5)$ you have $x-y \gt 0$ in the region above the diagonal and $x-y \lt 0$ in the region below. If you draw the other diagonal it cuts the square into a region where $x+y \gt 0$ and another where $x+y \lt 0$. If you break your integral ...


1

You have that $f_n \phi \to f \phi$ pointwise As $|f_n (x) \phi(x)| \leq |\phi(x)|$, $f_n \phi$ is dominated by $|\phi|$, that is an integrable fontion. Hence, all the hypothesis of the Lebesgue's Dominated Convergence Theorem are satisfied, and this imply that $$\lim_{n\to +\infty} \int_{\mathbb{R}} |f_n (x) \phi(x)-f(x)\phi(x)| dx = 0$$


1

When $n=m$ we have $\sin(nx)\cos(nx) = \frac{\sin(2nx)}{2}$. When $n\neq m$ use the product to sum formula: $$\sin(nx)\cos(mx) = \frac{\sin((n+m)x) + \sin((n-m)x)}{2}$$ Then the integration becomes much simpler.



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