New answers tagged

0

An adjunct to the asymptotic method of @George Simpson, eliminating the "it is possible to show" clause and avoiding the use of power series: We will show directly that for real $x\geq 0$ $$ 1-x^2 \leq e^{-x^2} \leq \frac{1}{1+x^2} $$ Let $a(x) = 1-x^2$ and $b(x)= e^{-x^2}$; then $a(0) = b(0) = 1$. And for $x\geq 0$, $$ \frac{da(x)}{dx} = -2x\\ ...


0

Note that $$\int_{0}^{1}\int_{0}^{1}\left(-\log\left(xy\right)\right)^{s}\left(\frac{1}{\log\left(xy\right)}+\frac{1}{1-xy}\right)dx $$ $$=-\int_{0}^{1}\int_{0}^{1}\left(-\log\left(xy\right)\right)^{s-1}dxdy+\int_{0}^{1}\int_{0}^{1}\frac{\left(-\log\left(xy\right)\right)^{s}}{1-xy}dxdy. $$ Now in the first integral take $x=e^{-u} $ and $y=e^{-v} $. We ...


1

By the way, $f(x)$ as shown is not continuous (but that makes no big difference). An alternative method (although the suggestion by @Thomas is the right way to approach such problems in general): $f(x)$ is piecewise linear on uniform intervals $(0,1), (1,2), \ldots (4,5)$. For any such function, breaking the function into trapezoids gives an exact result. ...


1

If we perform a change of variables $x=e^{-u},y=e^{-v}$ we may easily check that $$J(a,b) = \iint_{(0,1)^2}\frac{x^a y^b}{\log(xy)}\,dx\,dy = -\frac{1}{a-b}\,\log\left(\frac{1+a}{1+b}\right) \tag{1}$$ from which it follows that $$\begin{eqnarray*}K(a,b) &=& \iint_{(0,1)^2}\frac{x^a y^b(1-x)(1+y)}{\log(xy)}\,dx\,dy\\ &=& ...


3

The substitution is totally valid, no doubt about it. The answer that you have received on wolfram alpha has just differed by a constant from your result since it is an indefinite integral that we are talking about. If you carefully observe, the result you obtained on wolfram alpha can be expressed as: $$-\frac{1}{2}\cos t (\cos t+10)$$ ...


0

The integral is convergent by Dirichlet's test, since $\cos x$ has a bounded primitive while $\frac{1}{\sqrt[3]{\ln x}}$ is decreasing towards zero. Integration by parts gives: $$ \int_{e}^{M}\frac{\cos x}{\sqrt[3]{\log x}}\,dx = \left.\frac{\sin x}{\sqrt[3]{\ln x}}\right|_{e}^{M}+\int_{e}^{M}\frac{\sin x}{3x\ln x\sqrt[3]{\log x}}\,dx$$and $$ ...


1

HINTS: Find $f'(x)$: $$f'(x)=\frac{\text{d}f(x)}{\text{d}x}=\frac{\text{d}}{\text{d}x}\left(4\left(3-\cosh\left(\frac{x}{5}\right)\right)\right)=$$ $$4\left(\frac{\text{d}}{\text{d}x}\left(3\right)-\frac{\text{d}}{\text{d}x}\cosh\left(\frac{x}{5}\right)\right)=-4\left(\frac{\text{d}}{\text{d}x}\cosh\left(\frac{x}{5}\right)\right)$$ Now, use the chain ...


4

$$ \begin{align} \int_{-\infty}^{\infty} e^{-\frac{x^2} {2}} \Phi(x-t) dx & = \sqrt{2\pi}\int_{-\infty}^{\infty} \phi(x) \Phi(x-t) dx \\ & = \sqrt{2\pi}\Pr\{Z_1 \leq Z_2 - t\} \\ & = \sqrt{2\pi}\Pr\left\{\frac {Z_1 - Z_2} {\sqrt{2}} \leq - \frac {t} {\sqrt{2}}\right\} \\ & = \sqrt{2\pi}\Phi\left(- \frac {t} {\sqrt{2}}\right) \end{align} $$ ...


0

Old thread, but this question came up again and was marked a duplicate, so I have to post my physics lecture here. You know the symmetries of the problem that the electric field of a line charge must be purely radial and that its magnitude has no axial or azimuthal dependence $\vec E=E(r)\hat e_r$. Thus we may apply Gauss's law over a cylinder or radius ...


0

Let $\phi:[a,b]\rightarrow \Bbb R^2$ be some parametrisation of a curve $\Gamma$ with endpoints $\phi(a)=(1,2)=A$ and $\phi(b)=(3,4)=B$ To calculate the work done by the field $F$ as you move from $A$ to $B$ along $\Gamma$ you need to evaluate the line integral $\int_a^bF(\phi(t))\cdot \phi'(t) \,dt$ For example, you could choose your curve $\Gamma$ to be ...


-1

Hints: If $x\in [0,1]$, then $f(x)=\int_{-1}^{-x}-(x+t)dt+\int_{-x}^{0}(x+t)dt=\cdots$; If $x\in [1,2]$, then $f(x)=\int_{-1}^{0}(x+t)dt=\cdots$


3

Hint $$1 + \frac{x^4 - 8 x^2 + 16}{16 x^2} = \frac{16 x^2}{16 x^2} + \frac{x^4 - 8 x^2 + 16}{16 x^2} =\frac{x^4 + 8 x^2 + 16}{16 x^2}$$ Can you write the rightmost expression as a square of another expression?


5

Hint : You can notice that : $$\frac{1}{(x^2+2)^3}=\frac{1}{(2(\frac{x^2}{2}+1))^3}=\frac{1}{8((\frac{x}{\sqrt{2}})^2+1)^3}=\frac{\frac{1}{8}}{((\frac{x}{\sqrt{2}})^2+1)^3}$$ Now you should be able to use a result about $I_n$.


1

$\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{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} ...


0

I remember thinking about this problem in high school. Here is the intuitive explanation that popped into my head one morning before I got out of bed. I am assuming the question you are asking is why the antiderivative evaluated at the endpoints gives you the area under the curve. Let $f(x)$ be a function on an interval $[a,b]$. We want to calculate ...


1

$$\int \sum_{n=0}^\infty \frac{(-t^2)^n}{n!} dt$$ $$= \sum_{n=0}^\infty \int \frac{(-t^2)^n}{n!} dt$$ $$= \sum_{n=0}^\infty \int \frac{(-1)^n}{n!} t^{2n}dt$$ $$= \sum_{n=0}^\infty \left(\frac{(-1)^n}{n!} \int t^{2n}dt\right)$$ $$= \sum_{n=0}^\infty \left(\frac{(-1)^n}{n!} \frac{t^{2n+1}}{2n+1}\right)$$


2

Using integration by parts, the original integral turns into: $$ \int_{0}^{1}\frac{x-1}{1-x+\frac{x^2}{2}}\,\log(x)\,dx \tag{1}$$ We may compute the Taylor series of $\frac{x-1}{1-x+\frac{x^2}{2}}$ in a neighbourhood of the origin through partial fraction decomposition, then exploit $$ \int_{0}^{1} x^k\log(x)\,dx = -\frac{1}{(k+1)^2}.\tag{2} $$ That ...


1

Keep simplifying, we get $$ I = -\int_{0}^{\frac{1}{2}}\int_{0}^{\frac{\pi}{4}} r \cot\left(\frac{\theta}{2}\right) \log\left(1-r\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right)\right)\,d\theta\,dr$$ but $\cot\left(\frac{\theta}{2}\right)\sin\left(\theta+\frac{\pi}{4}\right) $ has a non-integrable singularity in a right neighbourhood of $\theta=0$, hence the ...


-3

For all smooth functions: $$f(x+\varepsilon)=f(x)+\varepsilon f'(x)$$ including $A(x)$ as the area function (for another function). So: $$A(x+\varepsilon)-A(x)=\varepsilon A'(x)$$ $$\varepsilon f(x)+\frac{1}{2}\varepsilon.\varepsilon f'(x)=\varepsilon A'(x)$$ where LHS1 is the rectangle area and LHS2 is the triangle area. LHS2 is now neglected: $$\varepsilon ...


1

You may use the Laplace transform: $$\mathcal{L}^{-1}\left(\frac{1}{x^6}\right)=\frac{s^5}{120},\qquad \mathcal{L}\left((x\cos x-\sin x)^3\right) = \frac{31104-196992 s^2-94080 s^4-13440 s^6}{(s^2+1)^4(s^2+9)^4}$$ turning the original integral into $$ \frac{1}{120}\int_{0}^{+\infty}\frac{31104s^5-196992 s^7-94080 s^9-13440 s^{11}}{(s^2+1)^4(s^2+9)^4}\,ds $$ ...


0

have you tried computing $\alpha (T) =e^{ik2\pi/T}$ and an integration by parts with $\alpha(T)^t$ ?


0

$\mathbf u = (1,-1), \mathbf v = (2,1)\\ (x,y) = (0,2) + \beta\mathbf u + \alpha\mathbf v\\ f(x,y) = y = 2 + \alpha - \beta\\ dx dy = ||\mathbf u \times \mathbf v||d\alpha\,d\beta = d\alpha\, d\beta\\ \int_0^1 \int_0^1 2 + \alpha - \beta \,d\alpha \,d\beta=2$


0

sorry my bad, -0.5*cos(2x) is the problematic function, which flips the sign for all the parts in the equation above which makes them different function which I think do not apply


1

All expressions differ only by a constant. That is why the derivative of each expression does agree with your initial function, since the derivative of a constant is zero. Another way of looking at these solution is that each solution is an indefinite integral of the given function.


1

An alternative approach to Marco Cantarini's perfectly sound answer. If we set, for any $\alpha>1$, $$ I(\alpha) = \int_{0}^{+\infty}\frac{e^{-x}-e^{-\alpha x}}{x}\,dx $$ differentation under the integral sign/Feynman's trick gives $$ I'(\alpha) = \int_{0}^{+\infty} e^{-\alpha x}\,dx = \frac{1}{\alpha}, $$ and since $\lim_{\alpha\to 1^+}I(\alpha) = 0$, ...


4

By Frullani's theorem we have $$\int_{0}^{\infty}\frac{e^{-x/\sqrt{3}}-e^{-x/\sqrt{2}}}{x}dx=\frac{1}{2}\log\left(\frac{3}{2}\right).$$


0

The periodic extension of $f(x)$ is $$ f_p(x)=\sum_{n\in\Bbb Z} f(x-2n\pi)=\arcsin(\cos x) $$ The function is even and then $a_0=0$ and $b_n=0$. \begin{align} a_n&=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos(nx)\,\mathrm dx\\ &=\frac{1}{\pi}\int_{-\pi}^0 \left(\frac{\pi}{2}+x\right)\cos(nx)\,\mathrm dx+\frac{1}{\pi}\int_{0}^\pi ...


0

I am not completely sure what your question is, so I'm sorry if I misunderstand you. You have to integrate over the entire period, i.e on the interval $[-\pi, \pi]$ (the function is $2\pi$ periodic and therefore the interval should have length $2\pi$). Hence you have that \begin{align} a_n &= \frac1\pi\int_{-\pi}^\pi f(x)\cos (n x)\,\mathrm{d}x\\ b_n ...


2

Hint: $0$ is not the problem for convergence. Consider what's happening at $1$ instead.


0

We may work with Lebesgue measure because $\int _{[a,b]}gdm=\int_{a}^{b}gdx.$ Suppose the claim is false, and assume, without loss of generality, that $\int gdm>0$. Now, $g$ is Riemann integrable, so it is continuous except perhaps on a set $Z\subset [a,b]$ such that $m(Z)=0$. Thus, $\int _{[a,b]}gdm=\int _{[a,b]\setminus Z}gdm>0$ so there is an ...


5

$$\int_{0}^{+\infty}\frac{dt}{\sqrt{(1+t^2)(1+x^2 t^2)}}=\frac{\pi}{2\,\text{agm}(1,x)}\tag{1}$$ hence $$ I(z)=\int_{0}^{1}\frac{x^z}{\text{agm}(1,x)}\,dx = \frac{2}{\pi}\int_{0}^{1}\int_{0}^{+\infty}\frac{x^z}{\sqrt{1+x^2\sinh^2(t)}}\,dt\,dx\tag{2}$$ that can be managed through integration by parts, getting the same recurrence relation of the Euler Beta ...


0

Need to show $$\lim_{x \rightarrow 0} \frac{R(x)}{x^3}=4! \lim_{x \rightarrow 0} x \int_0^{\infty} \frac{e^{-t}}{(1+xt)^5}\, dt=0$$ so we need to show $$ A= \lim_{x \rightarrow 0} x \int_0^{\infty} \frac{e^{-t}}{(1+xt)^5}\, dt=0 $$ Then $$0 \le A \le x \int_0^{\infty} {e^{-t}} \, dt = x $$ so $$0 \le A \le x$$ Then since $$\lim_{x \rightarrow 0^+}0 ...


-1

You can guarantee that an integral is 0 by three possible means. The function is zero on the entire interval a to b, except at discrete points that are unconnected. The function is balanced on the interval a to b such that the area below the axis is the area above the axis. The bounds of the integral are equal such that the area is over a region that is a ...


8

As $g$ is Riemann integrable, we know that for all$\epsilon>0$ there exists $\delta>0$ such that for all $n$ and partitions $a=x_0<x_1<\ldots < x_n=b$ with $x_{i}<x_{i-1}+\delta$ for $1\le i\le n$, and for all choices of $\xi_i\in[x_{i-1},x_i]$, $1\le i\le n$, we have $$\left|\int_a^b g(x)\,\mathrm dx ...


0

We set: $$\text{I}_y=\int\ln^y(x)\space\text{d}x$$ Now, by integration by parts: $$\text{I}_y=x\log^y(x)-y\int\log^{y-1}(x)\space\text{d}x=x\log^y(x)-y\cdot\text{I}_{y-1}$$ So, by induction we can find that: When $y=0$: $$\text{I}_0=x+\text{C}$$ When $y\ge1$: $$\text{I}_y=x\ln^y(x)-y\cdot\text{I}_{y-1}$$ Mathematica gives: ...


2

We can note that $$\int_{0}^{\infty}\frac{1}{1+\phi^{nx+1}}dx=\int_{0}^{\infty}\frac{\phi^{-nx-1}}{1+\phi^{-nx-1}}dx=\sum_{k\geq0}\left(-1\right)^{k}\phi^{-k-1}\int_{0}^{\infty}\phi^{-nx\left(k+1\right)}dx $$ $$=\sum_{k\geq0}\left(-1\right)^{k}\phi^{-k-1}\left[-\frac{\phi^{-nx\left(k+1\right)}}{n\left(k+1\right)\log\left(\phi\right)}\right]_{0}^{\infty} $$ ...


1

Hint: $\frac{dy^2}{dt}=2y\frac{dy}{dt}$, so $y^2(t)$ can also be written as $y^2(0) + 2 \int_0^t y(s) y'(s) ds$. What is $y'$ in this case?


1

Note that if $f$ would satisfy $f(a) = f(b) = 0$, you could take $g = f$ and get $\int_a^b f^2(x) \, dx = 0$ and since $f^2(x)$ is continuous and non-negative, you would immediately get $f(x) \equiv 0$. In general, you can modify $f$ continuously so that it stays the same on $[a + \frac{1}{n}, b - \frac{1}{n}]$ but drops down to zero afterwards. Define $$ ...


3

The proof starts like this: Suppose, by contraddiction that $ \exists x_0 \ \in \ (a,b) \ :\ f(x_0)>0 $ then $ \exists \delta>0 \ :\ f(x)>0 \ \ \forall x \in(x_0-\delta,x_0+\delta) $ Can you reach the contraddiction?


5

Your reasoning is wrong: the hypothesis asks that the integral be zero for all continuous $g: [a,b] \to \mathbb{R}$ with $g(a) = g(b) = 0$, not just for the particular choice $g = 0$. So $f = 1$ is invalidated because of the function $g$ which is a nonlinear quadratic which is $0$ at both $a$ and $b$.


3

This is to complete amd's comment. Compute the Jacobian of $u,v$ with respect to $x,y$: $$ \frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix} x & y \\ 2\,x & -2\,y\end{vmatrix}=-2(x^2+y^2)\implies \frac{\partial(x,y)}{\partial(u,v)}=-\frac{1}{2(x^2+y^2)}. $$ Now, when you do the change of variables, the Jacobian cancels the $x^2+y^2$ term.


2

One usually uses the notation $\int f(x) \, dx$ to denote an anti-derivative, or, more rigorously, the collection of all possible anti-derivatives of the function $f(x)$ over some domain that it usually understood implicitly from the context. Thus, one can interpret the statement $$ \int x^2 \, dx = \frac{x^3}{3} + C $$ as the statement that all the ...


4

You can consider $$ \int x(x^2-a)^2\,dx $$ as one of the functions $f(x)$ such that $f'(x)=x(x^2-a)^2$, which is nonetheless not determined until some further condition is imposed. You do have a further condition, that is, $f(a)=7$, so the function is determined. How do you determine it? You know that, for some constant $c$, $$ ...


0

$f(x) = \int g(x) dx$ is very bad notation. The $x$ in the integral sign is a so-called 'dummy variable'. $\int g(x)dx$ is just a number, it does not depend on $x$. The only thing it could possible mean is the function that has constant value $\int g(u)du$. When people erroneously write this down they usually mean $$f(x) = \int_a^x f(t)dt,$$ for some $a$. ...


0

Integrating we find \begin{align} a_n&= \frac{1}{\pi}\int_{0}^{\pi}\sin x \cos (nx) \,\mathrm dx\\ &=\frac{1}{2\pi}\int_{0}^{\pi}\left[ \sin ((n+1)x) - \sin ((n-1)x)\right]\,\mathrm dx\\ &=\frac{1}{2\pi}\left[\frac{\cos((n+1)x)}{n+1}-\frac{\cos((n-1)x)}{n-1} \right]_0^\pi\\ &=\frac{1}{2\pi}\left[\frac{-2 (n \sin x \sin(n x)+\cos x \cos(n ...


0

Probably it is easier to write $f(x)$ as $$ f(x)=\frac{1}{2}\left(\sin(x)+\left|\sin(x)\right|\right) $$ then recall that: $$ \left|\sin(x)\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1}.$$


0

Try solving, $$ \displaystyle\ \int { \dfrac{1}{(x^2+1)\sqrt{x^2-1}}} \mathrm{d}x $$ To know about the solution, visit my channel Calculus Society. I don't know what you would think about this, $$ \displaystyle\ \int { \dfrac{1}{\sec^2(x)+2\tan^2(x)}} \mathrm{d}x $$ or this, $$ \displaystyle\ \int { \dfrac{1}{1-\sin^4(x)}} \mathrm{d}x $$ Sorry I can't think ...


-1

Let $F(\alpha)=\int_0^2{f(x)f(\alpha+x)dx}$, changing in variable by letting $\alpha+x=u$, we get $F(\alpha)=\int_{\alpha}^{\alpha+2}{f(u-\alpha)f(u)du}$. Since $f$ is continuous then $F$ is differentiable and \begin{align*} F'(\alpha)=f(2)f(\alpha+2)-f(0)f(\alpha) \end{align*} but since $f(0)=f(2)$ and $f(x+2)=f(x)$, for all $x\in \mathbb{R}$. Then ...


5

Since $\;f(x)=2^x\;$ is a continuous function it is Riemann integrable in any bounded interval and thus you can choose that interval's partition as you like and also the points in each subinterval where the function is going to be evaluated. Thus, we can choose ...


2

By the triangle inequality, $$\int_{B_n} |f_n-f| \mathop{d\mu} \le \int_{B_n} (|f_n|+|f|) \mathop{d\mu} \le 2M \int_{B_n} \mathop{d\mu}$$ where the last inequality is due to $|f_n| \le M$ and $|f| \le M$.



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