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1

Try an easier $u$ sub. Hint: Try the one you used, but exclude any exponents.


5

You may just perform the change of variable $$v=\sin x, \qquad dv=\cos x\:dx,$$ giving $$ \int_{\pi/6}^{\pi/2} \frac{\cos(x)}{\sin^{5/7}(x)}\, dx=\int_{1/2}^{1} \frac{dv}{v^{5/7}}=\left[\frac72 v^{2/7}\right]_{1/2}^1=\frac{7}{2}\left(1-\frac{1}{2^{2/7}}\right). $$


1

Hint: $\cos x$ is the derivative of $\sin x$, and use the chain rule, where (the supersrcipt $n$ is the power, and $f'(x)$ is the derivative of $f(x)$ with respect to $x$) $$\frac{d}{dx}(f(x)^n)=nf(x)^{n-1}f'(x)$$


0

You certainly could use integration, but that would be tediously unnecessary. Thankfully, some simple geometry should suffice once you have some measurements: Find the area of the trapezoid face (left side on your picture) through the following formula: Then multiply that area by the "width" of the bin to find the total volume. Conceptually, ...


0

Setting $$ f(x)=\int_0^\pi\frac{\log(1+x\cos y)}{\cos y}\,dy, \quad x\in (-1,1), $$ we have $f(0)=0$, and $$ f'(x)=\int_0^\pi\frac{\cos y}{(1+x\cos y)\cos y}\,dy=\int_0^\pi\frac{1}{1+x\cos y}\,dy\quad \forall x\in (-1,1) $$ Setting $$ t=\tan\frac{y}{2}, $$ we have $$ y=2\arctan t,\, \frac{1}{1+x\cos ...


1

Let $x=\lambda^{-1/3} y$. Then the integral is $$\lambda^{-1/3} \int_0^{\infty} dy \, \exp{\left [-\lambda^{2/3} \left (y+\frac{4}{y^2} \right )\right ]} $$ By rescaling, we now may use Laplace's method, i.e., find the min of the exponential, Taylor expand about that min and evaluate the resulting integral. The min is at $1 - 8/y^3 = 0 \implies y=2$. The ...


0

Probably the most direct way to do this is by infinite series: expanding the logarithm, and interchanging the sum and integral, we have $$ I := \int_0^{\pi} \frac{\log{(1+x\cos{y})}}{\cos{y}} \, dy = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n} \int_0^{\pi} \cos^{n-1}{y} \, dy. $$ Firstly, note that $n-1$ has to be even for the integral to be nonzero, since ...


5

Note that $\frac{x^n}{x^n+1}=1-\frac{1}{x^n+1}$. Thus, $$\int_1^2 \frac{x^n}{x^n+1}dx=1-\int_1^2 \frac{1}{1+x^n}dx$$ For $n>1$, the integral on the right-hand side satisfies the inequality $$\left|\int_1^2 \frac{1}{1+x^n}dx\right|\le \int_1^2 x^{-n}dx=\frac{1-2^{1-n}}{n-1}$$ which clearly goes to zero as $n \to \infty$. Putting all of this together ...


0

Divide the numerator and denominator of the integrand by $x^n$: $$ \frac{1}{1+x^{-n}} $$ For every $x \in (1,2]$, $x^{-n} \to 0$ as $n \to \infty$. Since this is all but one point of the interval of integration, the integral tends to $$ \int_1^2 \frac{1}{1} \, dx = 1. $$ To do it more thoroughly than this, you can chop up the interval into a small region ...


7

Well, $\frac{x^n}{x^n+1} \leq 1$ for all $x \in [1,2]$, and the constant function $1$ is integrable in $[1,2]$, so by the Dominated Convergence Theorem we have: $$\lim_{n \to +\infty} \int_1^2\frac{x^n}{x^n+1}\,{\rm d}x = \int_1^2 \lim_{n \to +\infty} \frac{x^n}{x^n+1}\,{\rm d}x = \int_{1}^2 1\,{\rm d}x = 1.$$


0

The exponential term can be slightly rearranges as $$e^{-(1/\Delta t-i(E_0-E)/h)t}=e^{-(1/\Delta t)t}e^{i(E_0-E)/ht}$$ Taking the magnitude of the right-hand side and exploiting the fact that for real-valued $x$, $|e^{ix}|=|\cos x + i \sin x|=\sqrt{\cos^2x+\sin^2x}=1$, we find $$0\le |e^{-(1/\Delta t)t}e^{i(E_0-E)/ht}|\le e^{-(1/\Delta t)t}$$ and the ...


0

The idea is to use the identity $$ \cosh^2 u = \sinh^2 u + 1 $$ to simplify the square root; so, let $$ \sinh^2 u = 4y^2 \to \sinh u = 2y \to u = \sinh^{-1} 2y \\ \implies \cosh u du = 2dy \\ \sqrt {1 + 4y^2} = \cosh u \\ $$ from this you should be able to reoslve the integral.


-3

$$\int_0^\pi \frac{x}{(\sin x)^{\sin (\cos x)}}dx=$$ $$x\int_0^\pi sin^{-sin(cos(x))}(x) dx$$ I got no result in terms of standard mathematical functions!


-1

Hint: if you have $f(x)=F'(x)$, you can change the ''symbol'' for the variable, i.e write $t=x$, and you have: $f(t)=F'(t)$. Than: $$ \int_a^b f(x)=F(x)|_a^b=F(b)-F(a) $$ $$ \int_a^b f(t)=F(t)|_a^b=F(b)-F(a) $$ In your case you have: $$ \int_0^\pi\dfrac{\sin (t+\pi)}{t+\pi}dt=-\int_0^\pi\dfrac{\sin t}{t+\pi}dt=-\int_0^\pi\dfrac{\sin x}{x+\pi}dt $$


0

$\bf{My\; Solution::}$ Let $$\displaystyle I = \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx\;,$$ Let $\displaystyle x= \tan \phi\;,$ Then $dx = \sec^2 \phi.$ and Changing Limit, We Get $$\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^4 \phi\cdot \left(1-\tan \phi\right)^4d\phi.$$ $$\displaystyle =\int_{0}^{\frac{\pi}{4}}\tan^4 \phi \cdot \left(1-4\tan ...


1

For the function to have an extremum at a certain point $x_0$, it is not sufficient that $f'(x_0)=0$. What is sufficient is that $f'(x)$ alternates signs in a neighborhood of $x_0$, that is $f'(x)<0$ when $x_0-\epsilon<x<x_0$ and $f'(x)>0$ when $x_0<x<x_0+\epsilon$ or vice versa depending on the nature of the extremum. As you can see, ...


1

If $f$ is increasing then $$a_n=\sum_{k=1}^n\left( \underbrace{f(k)-\int_{k-1}^k f(x)dx}_{\ge0}\right)$$ so $(a_n)$ is increasing and $$f(k)-\int_{k-1}^k f(x)dx=\int_{k-1}^k (f(k)-f(x))dx\le f(k)-f(k-1)$$ so by telescoping we get $$a_n\le f(n)-f(0)$$ so if $f$ has a finite limit at $+\infty$ then $a_n$ is bounded above and then it's convergent.


0

Write $\frac{x}{x+1} = 1 - \frac{1}{x+1}$ and realize the problem reduces to $f=-1/(x+1)$. Now $$0 \leq -\frac{1}{1+k} + \int_{k-1}^k \frac{1}{1+x} \leq \frac{1}{k}-\frac{1}{k+1},$$ which can be shown by using the montonicity of $1/(1+x)$. But the above series is $O(1/k^2)$ and converges, so by the comparison test, the series we are interested in ...


1

Hint: $$ \frac 1n \cdot \left(f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+f\left(\frac{3}{n}\right)+ \ldots +f\left(\frac{n}{n}\right)\right) $$ is a Riemann sum for $\int_0^1 f(x) \, dx$ and the limit of $$ n \cdot \left(1-f\left(\frac{1}{\sqrt{n}}\right)\right) $$ can be calculated easily, for example with L'Hospital or using the Taylor series of ...


3

$$\left|\int_0^1 x^nf(x)\right| \le \int_0^1 |f(x)|x^n \le M\int_0^1 x^n = M \frac 1{n+1}$$ Where $M$ is the maximum of $|f(x)|$ in $[0,1]$ which exists thanks to the Weierstrass theorem. Now take the limit for $n \to \infty$ and use the squeeze theorem to conclude.


0

You can also define $y$ as a function of $x$ in the various intervals (it is simple since you have straight lines). For $0\leq x\leq 2$, $y=1+x$ For $2\leq x\leq 3$, $y=3$ For $3\leq x\leq 7$, $y=\frac{15}2-\frac{3}2x$ For $7\leq x\leq 9$, $y=-\frac{13}2+\frac{1}2x$ You know that $$\int_a^b(\alpha+\beta x)\,dx=\frac{1}{2} (b-a) \big(\beta (a+b)+2 \alpha ...


1

Some tricks to help you: a. count squares b. calculate the area of known figures, such as triangles and rectangles. For example, the integral in (b) can be calculate as the same of the area of a trapezoid($(1+3)\cdot 2/2$), a rectangle ($1\cdot 3$), and a triangle ($3\cdot 2/2$) to give a grand total of $10$. That should cover it, good luck.


-1

see the solution on your graph


4

No, this isn't true. As an example: $$f(x)=\begin{cases}1 &\text{if } x\in (0,1/2] \\ 0&\text{else}\end{cases}$$ and $$g(x)=\begin{cases}1 &\text{if } x\in (1/2,1) \\ 0&\text{else}\end{cases}$$


3

Counterexample: Consider $(0,1)$ with Lebesgue measure and the characteristic functions $\chi_{(0,\frac{1}{2})}$ and $\chi_{(\frac{1}{2},1)}$. Then $\chi_{(0,\frac{1}{2})}\chi_{(\frac{1}{2},1)} = 0$ but the product of their integrals is $\frac{1}{4}$.


2

$$\int_a^b{f(x)dx}=\int_a^{b}{f(x)[H(x-a)-H(x-b)]dx}=\int_0^{\infty}{f(x)[H(x-a)-H(x-b)]dx}$$


5

Others have focused on the specific value of the integral itself. However, it is clear from your description that you wish to evaluate something along the lines of $$ \frac{\int_0^1 f(x)r(x)dx}{\int_0^1 r(x)dx} $$ where $r(x)$ is the rational indicator function. While the component integrals cannot be evaluated, the overall concept is still meaningful. It ...


4

The integral exists in the Lebesgue sence and it is zero. That is because the rationals have measure zero, being a countable set. $r$ is the characteristic function of the rationals, and for such function the integral is given by $$ \int_0^1 r(x) \, dx = \lambda([0,1] \cap \mathbf Q) $$ where $\lambda$ is the Lebesgue measure. We have for measurable sets $A ...


15

It depends on whether you're considering the Riemann integral or Lebesgue integral. Note that every interval of any partition of $[0,1]$ contains a rational point, so any upper sum for $r(x)$ is $1$. Similarly, any lower sum for $r(x)$ is 0. Thus, $r(x)$ is not Riemann integrable. However, $r(x)$ is the indicator function for $\mathbb{Q}$, which is a ...


0

Observe that the quadratic $ x^2 -3x +3 $ has discriminant $$b^2-4ac=(-3)^2-4(1)(3)=9-12=-3$$ Hence no real solutions, which means it never crosses the $x$-axis, and since it opens up, this means it is always positive. Using this, and the following formula for $\arctan \frac{1}{u}$ $$\arctan \frac{1}{u}=\frac{1}{2}\pi - \arctan u, u>0 $$ We can rewrite ...


2

The crux to completing your work is to prove the following $$\int_0^1 \frac{\log^2(1-x)}{x^2} \,\mathrm{d}x = -2 \int_0^1 \frac{\log(1-x)}{x}\,\mathrm{d}x$$ You are close though, let me show you how to finish your work. I really liked this problem. Summarizing your work, you have proven that $$ \int_0^\infty \left(\frac{\log x}{1-x}\right)^2 ...


4

You can integrate by parts. A primitive to $1/(1-x)^2$ is $1/(1-x)-1=x/(1-x)$. The derivative of $(\ln x)^2$ is $2(\ln x)/x$. Thus $$ \begin{aligned} \int_0^1 \frac{(\ln x)^2}{(1-x)^2}\,dx &= \Bigl[\frac{x}{1-x}(\ln x)^2\Bigr]_0^1-\int_0^1 \frac{x}{1-x}\frac{2\ln x}{x}\,dx\\ &=-2\int_0^1\frac{\ln x}{1-x}\,dx. \end{aligned} $$


2

There is nothing inherently wrong with that, except you have to be specific about where $u$ lies. Using $ \sin{x} = \cos{(\pi/2-x)} $, we have $$ \cos{(\pi/2-x)} = \cos{u}, $$ so we can choose $u=\pi/2-x$. Going through with your substitution as-is requires some assumptions on the signs of the functions: $$ \cos{x} \, dx = -\sin{u} \, du, $$ and $$ \cos{x} ...


1

$f(x) \to \ell$, so for any $\varepsilon>0$ there is an $X_0>0$ such that $\lvert f(x)-\ell \rvert < \frac{1}{2}\varepsilon $. Then $$ \frac{1}{X}\int_0^X f(x) \, dx - \ell = \frac{1}{X}\int_0^X (f(x)-\ell) \, dx $$ Now apply the triangle inequality, $\lvert\int\rvert \leqslant \int \lvert \rvert$, and split the integral as the question suggests, so ...


0

$$f\;\;\text{is continuous}\implies\;\text{it has a primitive function, say}\;\;F\implies$$ $$\lim_{X\to\infty}\frac1X\int_0^X f(x)dx=\lim_{X\to\infty}\frac{F(X)-F(0)}X\stackrel{l'Hospital}=\lim_{X\to\infty}F'(X)=\lim_{X\to\infty}f(X)=L$$


0

The integration extends over singularities of the tangent function at $x=\pi/2$ and $x=3\pi/2$. To determine convergence, note that $$\int \tan xdx=-\log|\cos x|+C.$$ Therefore, the integral diverges logarithmically. However, if we interpret the integration in a Cauchy Principal Value sense, then we can write $$\begin{align} \int_0^{2\pi} \tan x ...


0

Hint: $$ \int \tan x dx=\int \dfrac{\sin x}{\cos x}dx = -\int(\cos x)^{-1} d(\cos x)=-\log(\cos x)) +C $$ so $$ \int_0^{\pi/2} \tan x dx = \lim_{x \rightarrow \pi/2}(-\log(\cos x))+1 $$ does not converge.


1

In general for an integral $\int_a^b f(x) dx$ to be well-defined, we need $\int_a^c f(x)dx$ and $\int_c^b f(x)dx$ both to be well-defined, for all $c \in (a,b)$. This is not the case for $\int_0^{2\pi}\tan(x)dx$ since $\int_0^{\pi/2}\tan(x)$ is not well-defined. However, the Cauchy principal value is $0$, since $$\int_{0}^{2\pi} \tan(x)dx = \lim_{\delta ...


1

This is an improper integral, and the usual way these are handled is by taking limits at each point of discontinuity (in this case, at $\pi/2$ and $3\pi/2$, from both sides at each). The result in this case is that the integral does not converge. There is good reason to do this, even if there is symmetry in the problem. For example, consider $\int_{-1}^1 ...


0

Maple produces a closed-form expression for concrete values of the parameters. For example, int(eval(x^(r/beta)*exp(-alpha*x)/(1-(1-beta)*exp(-x))^(alpha+1), [r = 2, beta = 1/2, alpha = 2]), x = 0 .. infinity); $$-48\,{\it polylog} \left( 4,1/2 \right) +42\,\zeta \left( 3 \right) - 4\,{\pi }^{2}\ln \left( 2 \right) +8\, \left( \ln \left( 2 \right) ...


1

As $$\frac{1}{(1-x)^s}=\sum_{j=0}^\infty{s+j-1\choose j}x^j ----(1)$$ for $|x|<1$ (reference is here). Since $e^{-x}$ lies between 0 and 1 for interval $(0,\infty)$. So we have two cases for $0<\beta<2$ and $\beta>2$. For $0<\beta<2$, using above result we can write $$\sum_{j=0}^\infty{\alpha+j\choose j}(1-\beta)^j\int_0^\infty ...


0

Your answer is correct but I used a different method. I substituted y for 3-x, evaluated the limits and calculated it from there.


0

Write $$ \frac{1}{ru^r} = \frac{1}{\Gamma(r+1)}\int_0^{\infty} \alpha^{r-1} e^{-u\alpha} \, d\alpha. $$ Then your integral becomes, after swapping the order of integration, $$ \frac{2}{\Gamma(r+1)} \int_0^{\infty} \alpha^{r-1} \left( \int_0^{\infty} e^{-u\alpha} \sin{u} \, du \right) \, d\alpha. $$ Now, we can do the internal integral by integration by ...


2

Write $$\int_0^{\infty} du \frac{\sin{u}}{u^r} = \operatorname{Im}{\left [ PV \int_0^{\infty} du \, u^{-r} e^{i u} \right ]} $$ Now consider the contour integral $$\oint_C dz \, z^{-r} e^{i z} $$ where $C$ is a 90-degree circular wedge of radius $R$ in the first quadrant of the complex plane, with a quarter circle of radius $\epsilon$ cut out at the ...


0

Hint: Use Euler's formula in conjunction with the well-known integral expression for the $\Gamma$ function, and then employ the reflection formula.


3

If $a=0$, then (assuming that $n\ge 1$) you always have $x^n = \int_0^x nt^{n-1}dt$. On the other hand, if $x^{n}=\int_a^x f(t)dt$ with $f$ being continuous. Take the limit of both sides $x\to a$ (they exist because the function are continuous). On the left side you obtain $a^n$, and the right side you obtain $0$, therefore necessarily $a=0$.


2

One approach is to integrate by parts twice $$ \begin{align} &\int_0^\infty e^{-\alpha x}\sin(\beta x)\,\mathrm{d}x\tag{1}\\ &=-\frac1\alpha\int_0^\infty\sin(\beta x)\,\mathrm{d}e^{-\alpha x}\tag{2}\\ &=\frac\beta\alpha\int_0^\infty e^{-\alpha x}\cos(\beta x)\,\mathrm{d}x\tag{3}\\ &=-\frac\beta{\alpha^2}\int_0^\infty\cos(\beta ...


2

Maybe the easiest (or at least most economic) way to solve this is using the fact that $\sin(\beta x)=\Im[{e^{i\beta x}}]$: $$ I=\Im\left[\int_0^{\infty}{e^{i\beta x- ax}}\right]=-\Im\left[\frac{1}{i\beta-a}\right]=\frac{\beta}{\beta^2+a^2} $$


1

We want to achieve $$U:=\iint_R x^2+2y^2 + 9 \, dx \, dy \ge \iint_R 2x^2+3y^2 \, dx \, dy=:V.$$ So $$\iint_R -(x^2+y^2)+9 \, dx \, dy \ge 0,$$ as you stated in your original post. For the double-integral to be positive, we should have $$-(x^2+y^2)+9 \ge 0,$$ like you said. Therefore, $$x^2+y^2 \le 9.$$ This is the equation of a circle of radius $3$, ...


3

$$ \begin{align} &\int_0^\pi\frac{x\sin(x)\,\mathrm{d}x}{1+\cos^2(x)}\tag{1}\\ &=\int_0^\pi\frac{(\pi-x)\sin(x)\,\mathrm{d}x}{1+\cos^2(x)}\tag{2}\\ &=\frac\pi2\int_0^\pi\frac{\sin(x)\,\mathrm{d}x}{1+\cos^2(x)}\tag{3}\\ &=\frac\pi2\left[-\arctan(\cos(x))\vphantom{\int}\right]_0^\pi\tag{4}\\[3pt] ...



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