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## New answers tagged integration

0

Start from the first Binet's formula: $$\ln\Gamma(z)=\left(z-\tfrac12\right)\ln z-z+\frac{\ln(2\pi)}2+\int_0^\infty\left(\frac12-\frac1t+\frac1{e^t-1}\right)\frac{e^{-t\,z}}t dt.\tag1$$ Change variable $t=-2\ln x$: $$\ln\Gamma(z)=\left(z-\tfrac12\right)\ln z-z+\frac{\ln(2\pi)}2+\frac12\int_0^1x^{2z-1}\frac{1-x^2+(1+x^2)\ln x}{(x^2-1)\ln^2x}dx.\tag2$$ ...

0

A classical method, going back to Newton himself, is to expand the integrand into a power series and integrate term by term. If the resulting series converges, it converges to the value of the integral (this is easy to show if the interval of integration is within the open interval of convergence; in your case, since $1$ is right at the edge of convergence, ...

5

$$\int_0^1e^{x-[x]}dx =\int_0^1e^{x }dx=e-1.$$

2

$$\int_0^{1000}e^{x-\lfloor x\rfloor}dx=\sum_{k=0}^{999}\int_k^{k+1}e^{x-\lfloor x\rfloor}dx=\sum_{k=0}^{999}\int_k^{k+1}e^{x-k}dx=\sum_{k=0}^{999}e^{x-k}\big|_k^{k+1}=\sum_{k=0}^{999}(e-1)=1000(e-1)$$

0

You question is not quite clear, since the full text of the question is not included. I think you are aked to find the integral from $0$ to $4$ of the function described by the picture. (The picture does not quite describe a function, because of the vertical line at $x=4$.) And probably you are also asked to find the integral from $0$ to $6$, or perhaps from ...

0

According to Gradshteyn and Ryzhik's "Tables of Integrals, Series, and Products", 1980 edition, page 595 formula 4.441 (2), there is a formula $$\int_0^{\pi/2} \ln \frac{p+q \sin ax}{p-q \sin ax} \frac{dx}{\sin ax}=\pi \arcsin \frac{q}{p},$$ with the restriction $p>q>0.$ The restriction is natural for the arcsine to exist on the right side, and maybe ...

2

Since $\log_2(x)=\frac{\ln x}{\ln 2}$ then your anti-derivative becomes $$\int\frac{\ln x}{x}dx=\int\frac1x\times\ln xdx=\frac12(\ln x)^2+C$$

0

The formula for Centroid $$f(x) = 9cos(x) ; g(x) = 9sin(x)$$ $$\bar x =\dfrac{ \int_0^{\frac{{\pi}}{4}} x(f(x)-g(x)) dx}{\int_0^{\frac{{\pi}}{4}} (f(x)-g(x))dx}$$ $$\bar y= \dfrac{ \int_0^{\frac{{\pi}}{4}} \frac{(f(x)+g(x))}{2}(f(x)-g(x)) dx}{\int_0^{\frac{{\pi}}{4}}(f(x)-g(x))dx}$$ To further elaborate: Denominator = $$9\int_0^{\frac{{\pi}}{4}}(cos(x) ... 0 Fix \alpha,\beta\in\mathbb{R}, then f(x)=(\alpha x+\beta)^2 attains its max at end points of [a,b] namely a or b. So (\alpha x+\beta)^2\leq \max\big\{(\alpha a+\beta)^2,\: (\alpha b+\beta)^2\big\}\leq (\alpha a+\beta)^2+(\alpha b+\beta)^2. Therefore \int_a^b (\alpha x+\beta)^2\leq(b-a)\Big((\alpha a+\beta)^2+(\alpha b+\beta)^2\Big) 0 It seems to be$$\frac{1}{4} \text{erf}\left(\frac{x}{\sqrt{2}}\right) \left(\text{erf}\left(\frac{a+\text{bx}}{\sqrt{2}}\right)+1\right)$$0 Hint: Given X a random variable with pdf f:$$\int_{-\infty}^\infty g(x)f(x)dx = \mathbb{E}\left[g(X) \right],\int_{a}^b g(x)f(x)dx = \int_{-\infty}^\infty g(x)I_{[a,b]}(x)f(x)dx = \mathbb{E}\left[g(X) I_{[a,b]}(X)\right],$$where indicator function I_{[a,b]}(x) is 1 if x\in [a,b], and 0 otherwise. On the sample space \Omega, ... 1$$ {\rm E}[g(X)\mathbf{1}_{0<X<a}] $$2 Sub x=a u; then$$\int_0^a dx \, \log{x} \, \log{(a-x)} = a \int_0^1 du \left [\log^2{a} + \log{a} \left (\log{u} + \log{(1-u)}\right ) + \log{u} \, \log{(1-u)}\right ] $$The first three integrals are straightforward; the middle two may be evaluated using the antiderivative$$\int dx \, \log{x} = x \log{x} - x +C$$For the final integral, you can ... 1 Here is another proof, the main part of which was communicated to me by Dr. Peter Otte of Bochum University: $$I_n := \int_{[0,1]^n}\mathrm{d}u\,\delta(1-\lvert u\rvert_1) \frac{1}{\prod_{j=1}^n (u_j + u_{j+1})} = (2\pi)^{n-2} \frac{[\Gamma(\frac{n}{2})]^2}{\Gamma(n)}.$$ First, define$$J_n(t) := ...

2

I rather dislike the way a lot of textbooks present the material on surfaces and solids of revolution because they will at some point present a table of formulas without having clarified the underlying reasoning. Students are confronted with several rather similar-looking equations and wonder how they are going to memorize them. What the surface area ...

1

In the proof that you cite there is an integral from $0$ to $\frac{\pi}{2}$, which for a circle means that it is in in the first quadrant (which makes sense, considering they were looking for the area of the quarter circle formed in the first quadrant), and because the $y$ values are positive, we use a positive square root. For a more graphic ...

0

The boundary surface, the sphere and the cone, intersect in a circle $(x^2+y^2 = \frac{1}{2})$, $z = \frac{1}{2}$. $z = r^2 and z = 1-r^2$ Equate these two to get r. The center of the sphere is $(0,0,0)$, so the upper part of the surface is exactly one quarter of the sphere of radius $\frac{1}{\sqrt{2}}$ as you are only interested in the upper part of the ...

1

Remember that $\sqrt{\phantom{x}}$ denotes the positive square root, so $$\sqrt{\cos^2\theta}=|\cos\theta|\ .$$ In the paper you linked, this occurs in an integral where $\theta$ goes from $0$ to $\pi/2$. For these $\theta$ values, $\cos\theta$ is positive, so $|\cos\theta|=\cos\theta$.

2

Substitute $x=\tan(\theta)$: \begin{align} \int_0^1\frac{\log(1+x^2)}{1+x^2}\,\mathrm{d}x &=2\int_0^{\pi/4}\log(\sec(\theta))\,\mathrm{d}\theta\\ &=-2\int_0^{\pi/4}\log(\cos(\theta))\,\mathrm{d}\theta\\ &=-\int_0^{\pi/4}\left[\log(1+e^{i2\theta})+\log(1+e^{-i2\theta})-2\log(2)\right]\,\mathrm{d}\theta\\ ... 1 With r constant, the differential element of area is r^{2}\sin\theta~ d\theta~ d\phi. The cone you have given is with \theta=45^{o}. Integrating as below $$\int^{\pi/4}_{\theta=0}\int^{2\pi}_{\phi=0} r^{2}\sin\theta~ d\theta~ d\phi=2\pi r^{2}(1-\cos{\pi/4})$$ 1 The parametric integrand you want is as follows. Putf(p)=\int_0^1\frac{x^p-1}{\ln x}dx$$Then$$f'(p) =\int_0^1\frac{\partial}{\partial p}\left(\frac{x^p-1}{\ln x}\right)dx=\int_0^1 x^p\;dx=\left.\frac{x^{p+1}}{p+1}\right|_0^1=\dfrac{1}{p+1}$$so f(p)= \ln(p+1)+ C, where C is to be determined. To find C, just note that f(0)=0 since the ... 3$$∫\dfrac{x^2}{(x+1)^3}dx$$Let u=x+1$$∫\dfrac{(u-1)^2}{(u)^3}du∫\dfrac{u^2-2u+1}{(u)^3}du∫\dfrac{1}{u}-\dfrac{2}{u^2}+\dfrac{1}{u^3}du$$I think you can take over from there. Integrate and back substitude u=x+1 NB:$$∫\dfrac{1}{u}du=ln(|u|)+c$$1 You don't have to use integration. This triangle is clearly a right triangle with side lengths 1 and 7. The centroid is the intersection of the three medians. Therefore, the slope of one median is the slope passing through the line (0, 7) and (0.5, 0) and the slope of another median is the slope of the line passing through (1, 0) and (0, 3.5). Saving you ... 1 As Andre Nicolas points out, the improper integral$$\int_0^{x^2}\frac{1}{t^3}\,dt$$diverges. If the lower limit were say, 1, then we could do the following calculation. Let u=x^2.$$\frac{d}{dx}\int_1^{x^2}\frac{1}{t^3}\,dt=\left(\frac{d}{du}\int_1^u\frac{1}{t^3}\,dt\right)\cdot \frac{du}{dx}=\frac{1}{u^3}(2x)=\frac{1}{(x^2)^3}(2x)=\frac{2}{x^5}$$0 When you want to check some process is a martingale. you need to check it is adapted ? it is integrable ? (since the conditional expectation is defined for integrable random variables[although you can define by approximation for nonnegative random variables of infinite expectation, if you want to.]) it has the martingale property ? In your solution, ... 2 Changing the variable$$t\mapsto -t$$implies the following change in the integral:$$ \int_{-\infty}^a\exp(-t^2)\,dt=-\int_{\infty}^{-a}\exp(-t^2)\,dt=\int_{-a}^\infty\exp(-t^2)\,dt. $$1 By expanding the (R-r)^2 piece, you see that you basically are looking to evaluate$$\int_s^R dr \frac{r^k}{\sqrt{r^2-s^2}}$$for k=1,2,3. For k=1, the integral is simple:$$\int_s^R dr \frac{r}{\sqrt{r^2-s^2}} = \frac12 \int_{s^2}^{R^2} \frac{du}{\sqrt{u-s^2}} = \sqrt{R^2-s^2}$$For k=2, integration by parts is useful:$$\begin{align}\int_s^R ...

2

Actually, from your definition $$\mathtt{f}[n]=\int_{n-\frac12}^{n+\frac12}f(x)\,\mathrm{d}x$$ and $$\mathrm{rect}(x)=\left\{\begin{array}{l} 1&\text{if }x\in\left[-\tfrac12,\tfrac12\right]\\ 0&\text{otherwise} \end{array}\right.$$ we have that for $n\in\mathbb{Z}$, \begin{align} \mathtt{f}[n] &=\mathrm{rect}\ast f(n)\\[9pt] ... 0 As far as I understand your notations,(comb \ast rect)(x) = 1,(f\ast rect(\cdot-n))(x) =\int_{n-1/2}^{n+1/2}f(x-y)dy, $$so the final convolution equals as a function of z$$\int_{\Bbb R}dx \int_{n-1/2}^{n+1/2}f(x-y)dy = \int_{\Bbb R}f(x)dx.$$3 This is a possible way (I assume a\geq 0, b\geq 0): 1) substitute x= a \sin t . 2) integrate the resulting integral by parts to get rid of \ln 3) you should end up with$$-\int_0^{\pi/2}\frac{2 a^2 b \sin^2 t\,dt }{b^2 -a^2 \cos^2 t} .$$4) it is possible to integrate the last integral by elementary means or using the residue theorem. You should ... 2 It is enough to compute \int \ln(a+\sqrt{1-x^2}) dx. Do a change of variables x=\sin t and integrate by parts:$$\int \ln (a+\cos t) \cos t \, dt = \ln(a+\cos t) \sin t + \int \frac{\sin^2 t}{a+\cos t} \, dt$$Use the substitution u = \tan t/2 so that \sin t = 2u/(1+u^2), \cos t = (1-u^2)/(1+u^2) and dt = 2/(1+u^2) du. This is known as the ... 7$$\begin{align*} I&=\int^1_0\frac{1-x^2+(1+x^2)\log x}{x+1}\frac{dx}{x\log^3x}\\ &=\left.\frac{-1}{2\log^2x}\frac{1-x^2+(1+x^2)\log x}{x+1}\right|^1_0-\int^1_0\frac{-1}{2\log^2x}\frac{\partial}{\partial x}\left(\frac{1-x^2+(1+x^2)\log x}{x+1}\right)dx\\ &=\int^1_0\frac{1}{2\log^2x}\frac{\partial}{\partial x}\left(1-x+\frac{(1+x^2)\log ...

0

Here is an idea I received from a friend. Use $\displaystyle 1/2\int_{0}^{2\pi}\left[log(a-e^{ix})+log(a-e^{-ix})\right]^{2}dx$ Then, use Gauss's Mean Value Theorem: $\displaystyle 2\pi f(a)=\int_{0}^{2\pi}f(a-e^{\pm ix})dx$ with $f(z)=log^{2}(z)$. This means that \displaystyle 1/2\int_{0}^{2\pi}log(a-e^{ix})dx+1/2\int_{0}^{2\pi}log(a-e^{-ix})dx=2\pi ... 2 Integration by parts twice yields \begin{align} \frac{\mathrm{d}}{\mathrm{d}a}\int_0^\infty\frac{1-e^{-at}}{t}\sin(t)\,\mathrm{d}t &=\int_0^\infty e^{-at}\sin(t)\,\mathrm{d}t\\ &=\frac1a\int_0^\infty e^{-at}\cos(t)\,\mathrm{d}t\\ &=\frac1{a^2}-\frac1{a^2}\int_0^\infty e^{at}\sin(t)\,\mathrm{d}t\\ &=\frac1{1+a^2} \end{align} Noting that ... 2 Since you know that $$\int_0^\infty \frac{\sin t}tdt=\frac\pi2$$ so it suffices to find $$\int_0^\infty\frac{e^{-t}}t\sin tdt$$ so let $$f(x)=\int_0^\infty\frac{e^{-t}}t\sin (xt)dt=\int_0^\infty h(x,t)dt$$ so using Leibniz theorem and since $$\left|\frac{\partial h}{\partial x}(x,t)\right|\le e^{-t}\in L^1((0,\infty))$$ so we have ... 6 Hints: One has\displaystyle\frac{1-\mathrm e^{-t}}t=\int_0^1\mathrm e^{-xt}\mathrm dx$For every real numbers$x$and$t$, one has$\mathrm e^{-xt}\sin t=\Im(\mathrm e^{-(x-\mathrm i)t})$For every complex number$z$such that$\Re z\gt0$, one has$\displaystyle\int_0^\infty\mathrm e^{-zt}\mathrm dt=\frac1z$For every real number$x$, one has ... 0 In spherical coordinates the intersection points$r=\sqrt 3/2$,$z=1/2$have colatitude$\varphi_0=\arctan\sqrt 3=\pi/3$and the second sphere is$\rho=2\cos\varphi: $$V= \int_0^{2\pi}\int_0^{\pi/3}\int_0^1\rho^2\sin\varphi d\rho d\varphi d\theta+ \int_0^{2\pi}\int_{\pi/3}^{\pi/2}\int_0^{2\cos\varphi}\rho^2\sin\varphi d\rho d\varphi ... 0 Answer using Cylindrical Coordinates: Volume of the Shared region = Equating both the equations for z, you get z = 1/2. Now substitute z = 1/2 in in one of the equations and you get r = \sqrt{\frac{3}{4}}. Now the sphere is shifted by 1 in the z-direction, Hence Volume of the Shared region =$$\int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{3}{4}}} ... 8 Here is a proof of Cleo's answer. Rewrite the integral as \begin{align*} I &= \int_0^1 \frac{\log(1-x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &= \int_0^1 \frac{\log(1-x^2)-\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \\ &= \int_0^1 \frac{\log(1-x^2)}{\sqrt{x}\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1+x)}{\sqrt{x}\sqrt{1-x^2}}dx \end{align*} The first integral can ... 1 I think the notationx^+=\max(x,0)$is nice. The function$F(x)=\sum_{n=1}^\infty (x-a_n)^+$is an antiderivative of your$f$, and therefore $$\int_a^b f(x)\,dx = \sum_{n=1}^\infty \bigg( (b-a_n)^+ - (a-a_n)^+ \bigg)$$ 5 The function$f(x)=a^x$for$a\geq 0$is uniquely determined by these three properties:$\forall x,y \in\Bbb R\left[f(x+y)=f(x)f(y)\right]f$is continuous at at least one point$f(1)=a$What Martín-Blas is trying to convey is that, once you agree that the fundamental property that an 'exponential-like' operation$E:\Bbb R\rightarrow \Bbb R$has is ... 0 For a sphere, we have$x^2+y^2+z^2=1 \Leftrightarrow z=\sqrt{1-x^2-y^2}f_y=-2y(\frac{1}{2})(1-x^2-y^2)^{-1/2}$,$f_x=-2x(\frac{1}{2})(1-x^2-y^2)^{-1/2}$. So by the formula for a surface area, we have $$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{(-2y(\frac{1}{2})(1-x^2-y^2)^{-1/2})^2+(-2x(\frac{1}{2})(1-x^2-y^2)^{-1/2})^2+1}dydx.$$ ... 1 Answer: The formula for Centroid $$\bar x =\dfrac{ \int_0^4 x(f(x)-g(x)) dx}{\int_0^4 (f(x)-g(x))dx}$$ $$f(x) = 5x^2+2x; g(x) = 0$$ $$bar x =\dfrac{\int_0^4 x(5x^2+2x) dx}{\int_0^4 (5x^2+2x)dx}$$ Rest is simple and if you evaluate, it will come to$2.9566$$$\bar y= \dfrac{ \int_0^4 \frac{(f(x)+g(x))}{2}(f(x)-g(x)) dx}{\int_0^4 (f(x)-g(x))dx}$$ $$\bar ... 1 You do this exactly the same way as your other surface area problem. Recall the surface area formula:$$f_y=2y, f_x=2x.$$So we have$$S=\int_{-5}^5\int_{-\sqrt{25-x^2}}^{\sqrt{25-x^2}}\sqrt{(2x)^2+(2y)^2+1}dydx$$Can you take it from there? Alternatively, you could do this using polar coordinates, which is much easier: ... 0 Recall the surface area formula: z=3-3y-5x, so f_y=-3, f_x=-5. So we have$$S=\int_{-4}^4\int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}}\sqrt{(-3)^2+(-5)^2+1}dydx=\int_{-4}^4\int_{-\sqrt{16-x^2}}^{\sqrt{16-x^2}}\sqrt{35}dydx=\int_{-4}^4\sqrt{35}(2\sqrt{16-x^2})dydx=2\sqrt{35}(\frac{1}{2}\sqrt{16-x^2}x+8sin^{-1}(\frac{x}{4}))|_{-4}^4dydx.$$... 1 \displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} y\;dz\;dy\;dx Taking the antiderivative in terms of dz, we have \displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} yz\;dy\;dx Now, evaluate from F(y) -f(0) \displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} yy\;dy\;dx \displaystyle\int_{0}^{1}\int_{0}^{1 - x^2}\int_{0}^{y} ... 7 As suggested in Greg Martin's comment, let us introduce F(x):=\int_x^1 f(t)dt, which satisfies that F'(x)=-f(x) and F(1)=0. Moreover, \int_0^1 F(x)dx=\int_0^1xf(x)dx=0, so by continuity, there exists x_0\in (0,1) such that F(x_0)=0. Let G(x):=e^{\int_0^xF(t)dt}F(x). Firstly, by definition, ... 1 It may help to write down the whole equation you have shown:$$ \int \sec^3 x \, dx = C + \sec x \tan x + \ln|\sec x \tan x| - \int \sec^3 x \ dx $$Now, what do you usually do when the thing you want to solve for appears multiple times in an equation? :) (I've added the "+C" because otherwise, the two copies of \int \sec^3 x \, dx might differ by a ... 2 If you did your algebra correctly, you're actually done. Let I=\int\sec^3(x)\,dx. Then you've demonstrated I=f(x)-I \implies I = f(x)/2, where f(x) is all that \sec{x}\tan{x} business you have. 2 Read the Wikipedia page at http://en.wikipedia.org/wiki/Integral_of_secant_cubed - this should give you your answer. You could continue with what you've done and use the integration constant C to get$$\int\sec^3(x)dx=\dfrac{1}{2}\left(C+\sec(x)\tan(x)+\ln|\sec(x)\tan(x)|\right)$\$

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