# Tag Info

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$$. These functions are not elementary, so they cannot be reduced to a finite combination of the arithmetic operations including exp and log (and the trig functions, but these are related to exp/log via the complex numbers). The best ... 2 With some help from a CAS I got this result with only one dilogarithm term: ... 0 A partial answer. For the first integral, perform the substitution \ln(\tan x)=-t. The integral is:$$-\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


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)}

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