Tag Info

New answers tagged

5

Notice that we can rewrite the limit as: $$ \left[\lim_{n\to\infty} \frac{\frac{\pi}{2n}}{\sin\left(\frac{\pi}{2n}\right)} \right]\left[\cos\left(\lim_{n\to\infty}\frac{\pi}{2n}\right)-\cos\left(\lim_{n\to\infty}\frac{(2n+1)\pi}{2n}\right) \right] = 1 \cdot [\cos 0 - \cos \pi] = 2 $$


2

Using $$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$ with $$\lim_{n\to\infty}\frac{\pi/(2n)}{\sin(\pi/(2n))}=1,$$ We have $$\frac{\pi/(2n)}{\sin(\pi/(2n))}\cdot (-2)\sin\left\{\frac{1+(2/n)}{2}\pi\right\}\sin\left(-\frac{\pi}{2}\right)\to 1\cdot (-2)\cdot 1\cdot (-1)=2\ (n\to \infty)$$


0

Hint: Let $u=\tan x$, multiply the whole series with $u^{2a}$, then differentiate with regard to u.


0

Set the center of the first circle in $(0,0)$ and that of the second circle in $(d,0)$. The equations of the circles are $y=\sqrt{r1^2- x^2}$ and $y=\sqrt{r2^2-(x-d)^2}$. In the segment connecting the two centers, the first and second circle intersect the x-axis in the points $(r1, 0)$ and $(d-r2, 0)$, respectively. From the two equations, we also get ...


3

We have the following closed form evaluation. $$ I:=\int_0^{\Large \frac{\pi}{4}}\!\!\left(\frac{1}{\ln(\tan x)}+\frac{1}{1-\tan x}\right)\! \mathrm dx= \color{blue}{\frac\pi8+\frac74\ln2+\frac\gamma2+\ln\pi-2\ln \Gamma\left(\frac14\right),} \tag1 $$ where $\gamma$ is the Euler-Mascheroni constant. A numerical approximation is $$ \color{blue}{I ...


0

This is for the definite integral of $x^{1/x}$, so it's not an answer perse. Compare with this. Note the wikipedia plate (from tetration) for the infinite exponential function: From calculus we have immediately: $$\int_a^b f^{-1}+\int_{f^{-1}(a)}^{f^{-1}(b)} f =bf^{-1}(b)-af^{-1}(a)$$ Now set $a=e^{-e}$ and $b=e^{1/e}$, with: ...


0

Related technique. You can use the power series approach. First note that $$ e^{x}-e^{-x}=\sum_{k=1}^{\infty}( 1-(-1)^k )\frac{x^k}{k!} = 2\sum_{k=0}^{\infty}\frac{x^{2k+1}}{(2k+1)!}.$$ Back to our integral we have $$ \int \frac{e^{x}-e^{-x}}{x}dx = \sum_{k=0}^{\infty}\frac{1}{(2k+1)!}\int x^{2k}dx + C = 2\,\rm Shi(x)+C . $$


1

The exponential integral function $\mathrm{Ei}(x)$ is defined by $$\mathrm{Ei}(x) = \int_{-\infty}^{x} \frac{e^{t}}{t} dt$$. These functions are not elementary, so they cannot be reduced to a finite combination of the arithmetic operations including exp and log (and the trig functions, but these are related to exp/log via the complex numbers). The best ...


2


0

A partial answer. For the first integral, perform the substitution $\ln(\tan x)=-t$. The integral is: $$-\int_0^{\infty} \frac{e^{-t}}{t(1+e^{-2t})}\,dt=-\sum_{k=0}^{\infty} (-1)^k\int_0^{\infty} t^{-1}e^{-(2k+1)t}\,dt$$ $$=-\lim_{a\rightarrow -1} \Gamma(a)\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^{a+1}}$$ I do not know how to evaluate the final limit.


3

There is no elementary anti-derivative for that function. Probably the most compact way to represent the integral in terms of special functions is in terms of the hyper-sine integral: $$\begin{align} \int\frac{e^x-e^{-x}}{x}\mathrm{d}x &=2\int\frac{\sinh{x}}{x}\mathrm{d}x\\ &=2\operatorname{Shi}{(x)}+\color{grey}{\text{constant}}. \end{align}$$


1

I know that upper limit is $\pi/2$, waiting for OP's clarification We have the following integral: $$I=\int_{0}^{\Large\frac{\pi}{2}}\left(\frac{1}{\log(\tan(x))}+\frac{1}{1-\tan(x)}\right)dx\tag{$I$}$$ Use $\displaystyle \int_0^a f(x) dx=\int_0^a f(a-x) dx$ ...


1

Allow me to present another approach. Though lengthier and more tedious, this method is independent of results derived from previous answers. The main idea is to expand ${\rm Li}_2(x)$ as a series and integrate term by term. For $n=1$, \begin{align} \mathcal{I}_1 &=\int^1_0{\rm Li}_2(x) \ {\rm d}x\\ &=\int^1_0\sum^\infty_{j=1}\frac{x^j}{j^2}{\rm ...


4

$$\int \frac{(x^4-4)dx}{(x^2\sqrt{4+x^2+x^4})}$$ $$ \int \frac{(x^2-4x^{-2})dx}{(\sqrt{4+x^2+x^4})} $$ $$ \int \frac{(x-4x^{-3})dx}{(\sqrt{4x^{-2}+1+x^2})} $$ Take, $$ u = 1+x^2+4x^{-2} \implies du = (2x-8x^{-3})dx \implies du/2 = (x-4x^{-3})dx$$ So integral = $$ \int \frac{du/2}{\sqrt{u}} = \sqrt{u} + C \implies \frac{\sqrt{4+x^4+x^2}}{x} +C $$


2

Physically, Hamiltonian operators in Quantum Mechanics should be semibounded, meaning that $(Ax,x) \ge M(x,x)$ for all $x\in\mathcal{D}(A)$ and for some fixed $M$. This has to be done with energy considerations. Second order ODES and PDES, in order to be symmetric, are quadratic in nature, and usually end up being semibounded--again, this is related to ...


0

If several change of variables are made, it can be shown that this integral is equivalent to the famous Vardi integral, $\displaystyle \int_{\pi/4}^{\pi/2}\ln(\ln(\tan(x)))dx=\frac{\pi}{2}\ln\left(\sqrt{2\pi}\frac{\Gamma(3/4)}{\Gamma(1/4)}\right)$, which has been done on the site. Well, it's twice the Vardi integral. Vardi's Integral: ...


2

As said in comments $e^{x^2}$ does not have an antiderivative expressible in terms of elementary functions. In fact, $$\displaystyle\int e^{x^2}\,dx=\frac{\sqrt{\pi }}{2} \text{erfi}(x)$$ in which appears the imaginary error function defined by $$\text{erfi}(x)=\frac{\text{erf(ix)}}{i}$$ with $$\text{erf(z)}=\frac{2}{\sqrt{\pi }}\int_0^ze^{-t^2}dt$$ ...


0

Re your first question, note that for every independent random variables $(X,Y)$, $X$ with density $f$, and every events $(A,B)$, $$P(X\in A,X+Y\in B)=\int_Af(x)P(Y+x\in B)\mathrm dx.$$ To show this, note that, in full generality, $$P(X\in A,X+Y\in B)=\iint \mathbf 1_{x\in A}\mathbf 1_{x+y\in B}\mathrm dP_{(X,Y)}(x,y),$$ hence, by the independence of ...


2

The Lerch transcendent, initially defined by $$\Phi(z,s,a):=\sum_{k=0}^\infty\frac{z^k}{(a+k)^s}, \quad a>0,\Re s>1,|z|<1,$$ admits the following integral representation $$ \Phi(z,s,a)=\int_0^{\infty}\frac{x^{s-1}e^{-ax}}{1-ze^{-x}}{\rm d}x. $$ By differentiation $$ ...


1

The region in $xyz$-space is the set of points which satisfy $x,y,z \ge 0$ and $\sqrt{x}+\sqrt{y}+\sqrt{z} \le 1$. Under the given transformation, the inequality $\sqrt{x}+\sqrt{y}+\sqrt{z} \le 1$ becomes $u+v+w \le 1$. Since we need our transformation to be 1:1, we need to restrict $u,v,w \ge 0$. The region in the $uvw$-space is the set of points ...


2

We can still do it directly. For the set up, it should be: $V = \displaystyle \int_{0}^1 \int_{0}^{(1-\sqrt{x})^2} \int_{0}^{(1-\sqrt{x}-\sqrt{y})^2} 1 dzdydx$


3

Integrate by parts twice. \begin{align} \int^1_0{\rm Li}_2^3(x){\rm d}x &=\left[x{\rm Li}^3_2(x)\right]^1_0+\int^1_03{\rm Li}^2_2(x)\ln(1-x){\rm d}x\\ &=\frac{\pi^6}{216}-\frac{\pi^4}{12}+6\color\red{\int^1_0\frac{\left[(x-1)\ln(1-x)-x\right]{\rm Li}_2(x)\ln(1-x)}{x}{\rm d}x}\\ ...


1

If you're looking for good numerical approximations, then for example $$\eqalign{- 0.000038839155&+ ( 2.472698342+ ( - 0.1185303776+ (\cr & 0.599575302+ ( - 4.237027543+ ( 9.643963778+ (\cr - & 12.22813840+ ( 8.085033824- 2.189508102\,x ) x ) x ) x ) x ) x ) x ) x} $$ is an optimal degree $8$ polynomial approximation on $[0,1]$, with maximum ...


2

We have that the function: $$ B(\lambda)=\sum_{n=0}^{+\infty}\left(\frac{1}{(2n-1)4^n}\binom{2n}{n}\right)^2\lambda^{2n} = \frac{1}{\pi}\int_{0}^{\pi}\sqrt{1+\lambda^2+2\lambda\cos(2\theta)}\,d\theta $$ satisfies the relation: $$B(\lambda) = 2\frac{1+\lambda}{\pi}\cdot E\left(\frac{4\lambda}{(1+\lambda)^2}\right)\tag{1}$$ and the differential equation: $$ B ...


1

Related problems and techniques: (I), (II). Here is a different form of solution $$ I = -3\sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{(-1)^k{ n\brack k}k(k-1) }{(n+1)^3n!} ,$$ where $ {n \brack k} $ is the Stirling numbers of the first kind.


3

Let's make a variable change $x=e^{-t}$: $$I=\int_0^{\infty}\frac{\sqrt{t}e^{-t}}{\sqrt{1-e^{-2t}}}dt$$ Now, let's expand the integrand into Taylor series: $$\frac{1}{\sqrt{1-e^{-2t}}}=\sum_{n=0}^{\infty}\frac{(2n)!}{(2^nn!)^2}e^{-2nt}$$ Thus $$I=\sum_{n=0}^{\infty}\frac{(2n)!}{(2^nn!)^2}\int_0^{\infty}\sqrt{t}e^{-(2n+1)t}dt=$$ ...


2

$\int_{-1}^{1}\left ( 8x^{3}+14x^{2}+6x+3 \right )dx$ $\left [ \frac{(8)(x^{4})}{4}+\frac{(14)(x^3)}{3}+\frac{6x^2}{2}+3x \right ]\bigg|_{-1}^{1} = \left [ 2x^4+\frac{14x^3}{3}+3x^2+3x \right ]\bigg|_{-1}^{1} = \left [ 2(1)^{4}+\frac{14(1)^{3}}{3}+3(1)^{2}+3(1) \right ] - \left [ 2(-1)^{4}+\frac{14(-1)^{3}}{3}+3(-1)^{2}+3(-1) \right ] = \left [ 2 + ...


0

$$\int_{-1}^1 (8x^3 + 14x^2 + 6x + 3)dx = \int_{-1}^1 8x^3 \,dx + \int_{-1}^{1} 14x^2\,dx + \int_{-1}^{1} 6x \,dx + \int_{-1}^1 3\,dx$$ Now use the power rule, and note that we can also pull out constant coefficients: $$\int_a^b cx^n \,dx= c\int_a^b x^n \,dx = c\cdot \frac{x^{n+1}}{n+1}\bigg|_a^b = c\left(\frac {b^{n+1}}{n+1} - \frac{a^{n+1}}{n+1}\right)$$


1

Hint: $\int_a^bf(x)dx = F(b)-F(a)$ where $F$ is any antiderivative of $F$.


2

Hints Use $\displaystyle \int x^n=\frac{1}{n+1}x^{n+1}$ Use the Fundamental Theorem of Calculus $\displaystyle \int_a^bf(x)\,\mathrm{d}x=F(b)-F(a)$


7

HINT : You can use $$\int_{\color{red}{-a}}^{\color{red}{a}}x^{2n-2}dx=2\int_{0}^{a}x^{2n-2}dx,\ \ \int_{\color{red}{-a}}^{\color{red}{a}}x^{2n-1}dx=0$$ where $n\in\mathbb N,a\not=0$. Hence, you can have $$\int_{-1}^{1}(8x^3+14x^2+6x+3)dx=2\int_{0}^{1}(14x^2+3)dx.$$ Then, $$\int x^mdx=\frac{1}{m+1}x^{m+1}+C$$ where $m\not=-1$.


5

HINT: $$\int_a^b x^n dx = \frac{x^{n+1}}{n+1}\bigg|_a^b$$ for $n\neq -1$.


3

$\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{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} ...


12

\begin{align} -\frac{1}{4}\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x &=-\frac{1}{4}\int^\infty_0\frac{\ln{\sinh{x}}}{1+\sinh{x}}e^{x/2}{\rm d}x\\ &=-\frac{1}{2}\int^\infty_1\frac{\ln\left(\frac{x^2-x^{-2}}{2}\right)}{1+\frac{x^2-x^{-2}}{2}}{\rm d}x\\ &=\int^1_0\frac{\ln(1-x^4)-\ln(2x^2)}{x^4-2x^2-1}{\rm d}x\\ ...


3

By "real random variable" you mean the values are real? That is, we have $P(X=\infty)=0$ ? Then sets $$ A_n = \{ X < n \} $$ have finite measure, and $$ \Omega = \bigcup_{n} A_n . $$


1

$$x^2 - y^2 + 2y = 1\iff x^2 = y^2 -2y+1 = (y-1)^2 \iff y-1 = \pm x \iff y = 1\pm x$$ Thus the two intersecting lines are given by $y = 1+x$ and $y = 1-x$. The point of intersection is given by $(0, 1)$. One bisector is the y-axis: the line $x=0$. The other bisector is given by $y = 1$. Can you take it from here?


3

Notice $$k\frac{dE(k) }{dk} = k\int_0^{\pi/2} \frac{\partial}{\partial k}\sqrt{1-k^2\sin^2\theta} d\theta = \int_0^{\pi/2} \frac{-2k^2\sin^2\theta}{2\sqrt{1-k^2\sin^2\theta}} d\theta = E(k) - K(k)$$ This leads to $$\frac{K(k)}{E(k)} = 1 - k\frac{d}{dk}\log E(k) = 1 - x\frac{d}{dx}\log E(k)$$ and hence $$\int\frac{dx}{x}\frac{K(k)}{E(k)} = ...


2

Your quantity is a Riemann sum for the integral $\int_0^{1/3} 3 f' \,dx$. By definition $3f$ is an antiderivative of $3f'$, so by the Fundamental Theorem of Calculus, this is $3[f(\frac{1}{3}) - f(0)]$. (Strictly speaking, the continuity of $f'$ ensures that we can apply FTC here.)


0

Note that : $$\frac{1}{h} \int_{-h}^h f(x) \ \mathrm{d}x =\frac{1}{ħ}\int_0^h f(x) \ \mathrm{d}x + \frac{1}{(-h)} \int_0^{(-h)} f(x) \ \mathrm{d}x$$ By the fundamental theorem of calculus ($f$ is continuous), the limit is $f(0)+f(0)=2f(0)$.


1

The question if weird, because the factorial function is defined for nonneative integers by $ n!=\prod_{k=1}^n k$ definition. That is why you rather could make a summation of it than an integral. The integration is not a tool for discrete functions. On the other hand as @L'universo said there is a generalization of factorial function called Gamma function ...


1

using the substitution $t=hu$ $$ \lim_{h\to 0} \frac{1}{h}\int_{-h}^hf(t)dt = \lim_{h\to 0} \int_{-1}^1f(hu)du = 2f(0) $$ since $f$ is continuous and bounded on $[-1,1]$


1

An other way to prove it: I note $F(x)=\int_0^x f(t)dt$ and I'll suppose that $h\to 0$. $$\frac{1}{h}\int_{-h}^h f(t)dt=\frac{F(h)-F(-h)}{h}=\frac{F(h)-F(0)}{h}-\frac{F(-h)-F(0)}{h}$$ Then $$\lim_{h\to 0}\frac{1}{h}\int_{-h}^hf(t)dt=\lim_{h\to 0}\frac{F(h)-F(0)}{h}-\frac{F(-h)-F(0)}{h}\underset{u=-h}{=}\lim_{h\to 0}\frac{F(h)-F(0)}{h}+\lim_{u\to ...


1

Assuming $\mathcal{C}$ are the continuous functions and $h\rightarrow 0$, we can use the mean value theorem, so there exists $x\in [-h,h]$ such that the integral is equal to $2hf(x)$. Again, using continuity of $f$, the expression converges to $2f(0)$.


2

If you use the gamma function definition of factorial $$x! = \int_0^\infty y^{x}e^{-y} dy$$ and change the order of integration in the resulting double integral $$\int_b^a x!dx = \int_b^a\int_0^\infty y^{x}e^{-y} dy dx = \int_0^\infty \int_b^a y^{x}e^{-y} dx dy $$ that might lead to a solution.


2

$$\int e^{2x} \sqrt{e^x+1}dx=\int (e^x\sqrt{e^x+1})e^xdx$$ $e^x+1=t,e^xdx=dt,e^x=t-1$ $$\int e^{2x}\sqrt{e^x+1}dx=\int(t-1)\sqrt tdt=\int t^{3/2}dt-\int t^{1/2}dt=$$ $$=\frac{t^{3/2+1}}{3/2+1}-\frac{t^{1/2+1}}{1/2+1}+C=\frac{2}{5}(e^x+1)^{5/2}-\frac{2}{3}(e^x+1)^{3/2}+C$$


1

HINT: $$\sqrt{e^x+1}=u\implies e^x+1=u^2$$


0

Rewrite the equation as $$y\frac{dy}{dx}=3x^2+4x^2.$$ Note that $$y\frac{dy}{dx}=\frac{d}{dx}\left(\frac{y^2}{2}\right),$$ by the Chain Rule. So the derivative of $\frac{y^2}{2}$ is $3x^2+4x$. It follows that $$\frac{y^2}{2}=x^3+2x^2+C,$$ for some constant $C$. Use the value of $y$ when $x=1$ to evaluate $C$. Now we know everything.


0

$\dfrac{dy}{dx}=\dfrac{3x^2+4x}{y}$ Now cross-multiply to get: $y\,\mathrm{d}y=3x^2+4x\,\mathrm{d}x$. Integrate on both sides to get: $\frac{1}{2}y^2=x^3+2x^2+C$. Multiply by two on both sides to get: $y^2=2x^3+4x^2+C$. Now solve for $C$ by plugging in the coordinate pair $(1, \sqrt{10})$: $10=2\cdot1^3+4\cdot1^2+C$ gives $C=10-2-4=4$. $y^2$ becomes: ...


0

Rearranging: $$ydy=(3x^2+4x)dx$$ Integrate: $$\int ydy=\int(3x^2+4x)dx$$ $$\frac{y^2}2=x^3+2x^2+C$$ Put $(1,rad10)$ to get the value of C, then put $x=0$ to get y


0

$$\frac{\tan\left(\dfrac\pi4 + t\right) - \tan\dfrac\pi4}t =\frac{\sin\left(\dfrac\pi4 + t-\dfrac\pi4\right)}{t\cos\left(\dfrac\pi4 + t\right)\cos\dfrac\pi4}$$ $$\implies\lim_{t\to0}\frac{\tan\left(\dfrac\pi4 + t\right) - \tan\dfrac\pi4}t=\frac1{\cos\dfrac\pi4}\cdot\lim_{t\to0}\frac{\sin t}t\cdot\frac1{\lim_{t\to0}\cos\left(\dfrac\pi4 + t\right)}$$



Top 50 recent answers are included