# Tag Info

5

Set $t=\frac{e^x}{4}$ , or $e^x=4t$ , thus $4dt=e^x dx$ as $x\to-\infty$ then $t\to 0$ as $x\to\ln(4)$ then $t\to 1$ we have $$I=\int_{-\infty}^{\ln(4)}\frac{xe^x}{\sqrt{4e^x-e^{2x}}}dx=\int_{0}^{1}\frac{\ln(4t)}{\sqrt{16t-16t^2}}4dt=\int_{0}^{1}\frac{\ln(4)+\ln(t)}{\sqrt{t-t^2}}dt$$ I=\ln(4)\int_{0}^{1}\frac{1}{\sqrt{t-t^2}}+\int_{0}^{1}\frac{\ln(t)}{... 2 \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}} \... 1 Here is a slightly different variation to OPs example, followed by another one. Suppose p is an even function, i.e. p(x)=p(-x) and q(x)q(-x)=1. Then \begin{align*} \int_{-a}^{a}\frac{p(x)}{1+q(x)}\,dx=\int_{0}^{a}p(x)\,dx \end{align*} A proof of the statement together with an application can be found in this answer. Note: This technique ... 4 We can write the interal in (4) asI=-\int_{0}^{1}\frac{1-x+\log\left(x\right)}{\log^{2}\left(x\right)}dx $$now define$$I\left(\alpha\right)=-\int_{0}^{1}\frac{x^{\alpha}\left(1-x+\log\left(x\right)\right)}{\log^{2}\left(x\right)}dx,\,\alpha\geq0 . $$We have$$I''\left(\alpha\right)=-\int_{0}^{1}x^{\alpha}\left(1-x+\log\left(x\right)\right)dx=-\...

4

Through the substitution $x=e^{-t}$ the original integral equals: $$I=\int_{0}^{+\infty}\left(2\frac{e^{-t}-1}{t^2}+\frac{e^{-t}+1}{t}\right)e^{-t}\,dt \\=2\color{purple}{\int_{0}^{+\infty}\frac{e^{-t}-1+t}{t^2}\,e^{-t}\,dt}+\color{blue}{\int_{0}^{+\infty}\frac{e^{-t}-1}{t}\,e^{-t}\,dt}$$ where the blue integral is yet manageable through Frullani's theorem (...

1

The total derivative of $f\left(x(t),y(t)\right)$ is : $$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$ Considering now the same function but depending on a parameter $a$, say $f_a\left(x(t),y(t)\right)$ : If $a$ isn't function of $t$ and if $t$ is the only variable the total derivative is the same : $... 1 Any function$g(x)$such that$g(c+a)+g(c-a)=k$for all$a$on the interval$(0,b)$with any function$f(x)$such that$f(c-a)=f(c+a)$for all$a$on the interval$(0,b)$will satisfy the equation $$\int_{c-b}^{c+b}{f(x)g(x)dx}=k*\int_{c}^{c+b}{f(x)dx}$$ because, using a trapezoidal Riemann sum after splitting the integrals into $$\int_{c-b}^{c}{f(x)g(x)dx}... 1 \newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 0 \newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 0 Ahmed Hussein's deleted answer is basically correct: near x=0 we have 1-\cos(x)^a \sim \frac a2 x^2, and so the integral converges or diverges along with \int_0 \frac1{x^{b+2}}\,dx, which is to say, converges when b<3 and diverges when b\ge3. The integral isn't improper at x=\frac\pi2, so nothing needs checking there. (The question specifies ... 1 It is not referring to the area you show. By definition of "area enclosed" in polar coordinates, it should be the area enclosed for the same range of \theta, in other word I think the author is referring to the area on the right hand side:$$A=\int_{-\pi/2}^{\pi/2}\left[\frac{1}{2}(2(1+\cos\theta))^2-\frac{1}{2}2^2\right] d\theta=\int_{-\pi/2}^{\pi/2}... 2 Given the formula for the area in polar coordinates and the symmetry of the configuration, the area you want to compute is just: $$2\pi + \int_{\pi/2}^{\pi}\left[2\left(1+\cos\theta\right)\right]^2\,d\theta =\color{red}{5\pi-8}.$$ 3 Consider the integral $$I= - \int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} (\sin^{-1} x)^4 \,dx.$$ Since $$(\sin^{-1} x)^4 = \frac32 \sum_{n=1}^{\infty} \cfrac{2^{2n} H_{n-1}^{(2)}}{n^2 \binom{2n}{n}} \,x^{2 n} \tag{1}$$ and $$-\int_0^1 \frac{\ln(1-x^2)}{\sqrt{1-x^2}} x^{2n}\,dx= \frac{\pi}{2} \binom{2n}{n} \frac{(H_n + 2\ln2)}{2^{2n}}, \tag{2}$$ we have $$\... 2 Because i is a constant with respect to the variable of integration, we can use the fact that for any constant b:$$\int \frac{b}{(b+x^2)^{3/2}}dx = \frac{x}{\sqrt{x^2 + b}}$$so that$$\int_0^i \frac{i}{(i+x^2)^{3/2}dx} = \frac{i}{\sqrt{i^2 + i}} = \sqrt{\frac{i^2}{i^2+1}}.$$When we square this (to get the ith term in the sum), we get:$$\left(\... 3 This really isn't so bad. $$\int_0^i\frac{i}{(i+x^2)^{3/2}}dx=\frac{i}{\sqrt{i(i+1)}}.$$ So you're summing: $$\sum_{i=m}^n\frac{i}{i+1}.$$ The latter sum unfortunately doesn't have an explicit form, unless you're willing to use digamma functions. 0 Let us define: $$I_3(a) := \int\limits_{[0,1]^3} \sqrt{x+\sqrt{y+\sqrt{z+a}}} dx dy dz$$ Then by using elementary integration we have the following result: \begin{eqnarray} I_3(a) &=& \frac{32}{31} \left( \sum\limits_{k=0}^3 |[\begin{array}{r} 3 \\ k \end{array}]|(-1)^{3-k} \frac{(u^+)^{\frac{9}{2}+k} - (u^-)^{\frac{9}{2}+... 1 Almost, but not quite. I'm seeing a lot of notational issues and other issues around here which could lead to trouble. Instead, say:$u = f(x), \qquad dv = g(x) \, dxdu = f'(x)\, dx, \qquad v = \displaystyle \int_0^x g(t) \, dt + v(0)\qquad $(this$v$is explained later in the post) Then we have $$\int_0^a f(x) g(x) \, dx = \left[f(x) \left(\int_0^... 7 It's good, if read correctly; you are less likely to make errors with this if you set$$ G(x)=\int_0^xg(t)\,dt $$and write$$ \int_0^af(x)g(x)\,dx= \Bigl[f(x)G(x)\Bigr]_0^a-\int_0^a f'(x)G(x)\,dx $$If you prefer not to use G, you can write$$ \int_0^af(x)g(x)\,dx= \left[f(x)\int_0^x g(t)\,dt\right] - \int_0^a f'(x)\left(\int_0^x g(t)\,dt\right)\,dx $$2 Yes!It is correct Let u=f(x), dv=g(x)dx, du=f'(x)dx, v=\int_{0}^{a}g(x)dx so \int_{0}^{a}f(x)g(x)dx=[f(x)\int_{0}^{x}g(x)dx]_{0}^{a}-\int_{0}^{a}[\int_{0}^{x}g(x)dx]f'(x)dx Next, integrate by parts a second time, u=f'(x)dx, dv=\int_{0}^{a}g(x)dx du=f''(x)dx, v=\int_{0}^{a}\int_{0}^{a}g(x)(dx)^{2} \int_{0}^{a}f(x)g(x)dx=[f(x)\int_{0}^{x}g(... 0 Hint: With a linear transform of u, you can decompose as two indefinite integrals of the form$$\int\frac{u}{u^2+c^2}e^{-u^2}du=\frac12\int e^{-u^2}d\ln(u^2+c^2)=\frac{e^{c^2}}{2}\int e^{-e^{t^2}}dt.$$Because of the double exponential, you shouldn't expect a closed form, but thanks to the very fast decreasing behavior, numerical integration over a ... 1 Just an addendum, maybe some of this will be useful to you:$$\int_0^1 \frac{(1-x) \log (1-x)}{x \log x} \, {\rm d}x==1.2577468869\dots= \gamma_{1}(1,0) - \gamma=\int_0^1 \ln (1-x) \ln \left(\ln \left(\frac{1}{x}\right)\right) \, dx-\gamma==\sum_{k=1}^{\infty} \frac{\ln (k+1)}{k(k+1)}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\sum_{k = 2}... 0 Concerning the special case where$y=0$the problem is $$I=\int_0^\infty \frac{e^{-\frac{(u-x)^2}{2 A}}-e^{-\frac{(u+x)^2}{2 A}}}{\sqrt{2 \pi } \sqrt{A}x \,u }\,du$$ Integration leads to $$I=\frac{e^{-\frac{x^2}{2 A}} \left(\log \left(\frac{x}{A}\right)-\log \left(-\frac{x}{A}\right)\right)}{\sqrt{2 \pi } \sqrt{A} x}$$ provided$\Re(A)\geq 0. Now $$\... 1 (1) Note that your integrand is periodic in all the integration variables. We will use this fact later. Let us introduce the new variables x_1= \phi_1- \phi_2, x_2 = \phi_2 -\phi_3 and x_3 = \phi_3-\phi_1. We immediately notice that x_3 = - x_1 - x_2. The goal is thus, to perform a change of variables from (\phi_1, \phi_2 , \phi_3) to (X, x_1, ... 1 \newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 3 It's worth getting some intuition for problems of this kind. The fact that the function is nonnegative means if you plot the function then it doesn't dip below the x-axis. If it is continuous, then very roughly speaking, you can plot the function without taking your hand off the page. Finally, if the definite integral is zero then the area under the curve ... 1 Suppose that there exists x_0 s.t f(x_0) > 0 . f is continuous so there exists \delta >0 s.t for each x if x\in[x_0-\delta,x_0+\delta] than f(x) > \dfrac {f(x_0)}{2} > 0 (for \epsilon = (\dfrac {f(x_0)}{2} ) now split the integral of [a,b] to [a,x_0-\delta],[x_0-\delta,x_0 +\delta] ,[x_0+\delta, b] and get to a ... 3 Suppose that f(z)> 0 for some z\in[a,b]. Then, since f is continuous, there exists a neighbourhood (c,d) around z such that f(x)>\frac{f(z)}{2}>0 for all x\in(c,d). But then \int_c^df(x)dx>0. You should be able to complete the proof from there. 0 Though it is a bit roundabout, you can define \ln(x)=\int_1^x \frac{1}{t} dt and \exp to be the inverse of \ln. It then follows that \exp(1) is your e. You can now develop the Taylor series of \exp. From this development you find e=\sum_{n=0}^\infty \frac{1}{n!} and also$$0<e-\sum_{n=0}^N \frac{1}{n!}<\frac{e}{(N+1)!}.$$This follows ... 3 Hint. An elementary approach. One may consider$$ I_0=1-\frac1{e}, \quad I_n=\int_{\large\frac1{e}}^1\left(-\ln x \right)^n\:dx, \quad n\ge1, $$then integrating by parts,$$ \begin{align} I_n&=\left[ x \frac{}{}\left(-\ln x \right)^n\right]_{\large\frac1{e}}^1 +n\int_{\large\frac1{e}}^1\left(-\ln x \right)^{n-1}\:dx \\&=-\frac1{e}+nI_{n-1} \end{... 0 Ifx\in [0;\infty)$, then$x+1\in [1;\infty)$So by substitution:$u\in[1;\infty)$Hence: $$\int_0^\infty g(x+1)\operatorname d x = \int_1^\infty g(u)\operatorname d u$$$\blacksquare$3 The original integral has$x$ranging from$0$to$\infty$. After the substitution these need to be translated to bounds for$u$. Since$u=1+x$, the lower bound$x=0$corresponds to$u=1$. -1 Notice for$0 \le x \le 3$then$x - 3 \le 0$so$\sqrt{x - 3} = \sqrt{-1*(3 - x)} = i \sqrt{3 - x}$is strictly$0$or purely imaginary. So$\int_{0}^3 \frac 1{\sqrt{x - 3}} dx = \int_{0}^3 \frac 1{i\sqrt{3-x}} dx = \frac 1i \int_{0}^3 \frac 1{\sqrt{3-x}} dx$. Now$\int_{0}^3 \frac 1{\sqrt{3-x}} dx= - \sqrt{3-x}|_{0}^3= \sqrt{3} $is purely real,$\frac ...

-1

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 4 The connection to hyperbolic functions is due the following Fourier series, valid when$|z|<\pi$: $$\frac{\pi}{x} \frac{\cosh z x}{\sinh \pi x} = \sum_{n \in \mathbb{Z}}\frac{(-1)^n \cos( z n)}{n^2+x^2} .\tag{1}$$ It is obtained quite straightforwardly. For instance, see @Machinato's related calculation in this answer, or the calculation on this page. ... 4 The truth is the concept in your mind (i.e., geometric area and integration) applies to real scalar functions with single variable, while you're not integrating a real function. In your case, assuming the integrand function ($1/\sqrt{x-3}$) to be a real scalar function, it is not defined over the integration interval$(0,3). But, if you assume it to be a ... 5 The area under the curve idea applies to real-valued functions. This function is imaginary for almost all values of the area of integration. 2 We may exploit Frullani's theorem to get an integral representation of our series. $$\begin{eqnarray*}S=\sum_{n\geq 1}\frac{\log(n+1)-\log(n)}{n}&=&\int_{0}^{+\infty}\sum_{n\geq 1}\frac{e^{-nx}-e^{-(n+1)x}}{nx}\,dx\\ &=&\int_{0}^{+\infty}\frac{1-e^{-x}}{x}\left(-\log(1-e^{-x})\right)\,dx\\&=&\int_{0}^{1}\frac{x\log x}{(1-x)\log(1-x)}\... 2 The given series admits a closed-form in terms of the poly-Stieltjes constants. The poly-Stieltjes constants arise in the context of finding the Laurent series expansion of the poly-Hurwitz zeta function$$ \begin{align} \zeta(s\mid a,b)= \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \tag1 \end{align} $$around s = 0. One may prove that (see Theorem ... 2 \newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 1 Hint. From standard properties (see this table) and from the given result one gets$$ \mathscr{L}\left(t \sin bt \right)(s)=-\left(\mathscr{L}\left( \sin bt \right)\right)'(s)=\frac{2bs}{(s^2+b^2)^2} $$then using$$ \mathscr{L}\left(e^{at}f(t) \right)(s)=\left(\mathscr{L}f\right)(s-a) $$gives finally$$ \mathscr{L}\left(t e^{at}\sin (bt) \right)(s)=\... 0 We can use the identity $$\mathrm{cl}_2(x)^2 = \mathrm{cl}_2(2\pi -x)^2$$ Hence we have $$I(5)=\int_0^{2\pi}\operatorname{Cl}_2(x)^2\,(2\pi -x)^5\,dx$$ Expand $$I(5) = 32 \pi^5I(0)-80 \pi^4 I(1)+80 \pi^3 I(2)-40 \pi^2 I(3)+10 \pi I(4)-I(5)$$ Solve forI(5)$$$I(5) = 16\pi^5I(0)-40\pi^4 I(1)+40\pi^3 I(2)-20\pi^2 I(3)+5\pi I(4)$$ This simplifies ... 1 You are right. This question is nonsense. The only finite areas in the diagram are: The area between the curve and the line$y=x+3$. This is is divided in two parts (also finite) by the line$x=0$. But the line$x=3$is irrelevant. The area at the top right of your picture, above$y=x+3$, to the left of$x=5$, and to the right of$y=x^2+9$. But the line$x=...

1

Hint How do you define the set of points in the blue region in terms of inequalities? $$\text{BlueRegion}=\{(x,y)|0 \le x \le 6 , 0 \le y \le \color{red}{\text{?}}\}$$ There is also another way to do it $$\text{BlueRegion}=\{(x,y)|0 \le y \le \frac 23 , 0 \le x \le \text{?}\}$$

1

Your intuition is very true. You are adding apples up. For a non-wild function $\int_a^b f(x)dx$ means adding up (infinitely many) values $f(x)$. But why does it become finite? Because each value $f(x)$ only occupies the single point $x$ on the interval $[a,b]$. So, it is adding up very many values but each value being multiplied by an almost zero segment. ...

4

I think that your confusion stems from a misunderstanding of how summations relate to integrals. Think of an integral as the limit of a Riemann Sum, or $Lim_{\Delta x\rightarrow 0} \Sigma f(x)\Delta x$. Let's think for a second about how we derive this. We know an integral is the area under a curve. But how can we calculate this without calculus? Well, we ...

1

Assuming it is $\;y=\frac19x\;,\;\;y=0\;,\;\;x=6\;$ , the integral can be put in the form $$\int_0^6\int_0^{\frac19x}(x+y)dydx$$

0

You start with the inner integral: $\int_0^{1/9x}(x+y)dy=[xy+1/2y^2]_0^{1/9x}=10/9x+1/162x^2$ and this function you integrate with dx on the limits of $0,6$.

1

Assuming that $d$ and $n$ are positive integers, you can just use the binomial theorem to expand the integrand, and integrate term by term, since it's a polynomial in $y$. $$(1 - u)^n = \sum_{i = 0}^n (1)^{n-i}{n \choose i} (-u)^i = \sum_{i = 0}^n (-1)^i{n \choose i} u^i$$ In your case, $u = y^d$, so this becomes $$(1 - y^d)^n = \sum_{i = 0}^n (-1)^i{n \... 8 Here is yet another approach. We first note that we can write \frac{x-1}{\log(x)} as$$\frac{x-1}{\log(x)}=\int_0^1 x^t\,dt$$Therefore, we can write$$\begin{align} \int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\int_0^\infty \int_0^1 \frac{x^t-1}{\log(x)}\,\frac{1}{1+x^2}\,dt\,dx\\\\ &=\int_0^1 \int_0^\...

0

$e^{ix}$ is a vector of length 1 and argument $x$. $e^{ix}dx$ is a vector of length $dx$ and argument $x$. The sum of all these vectors is, intuitivelly speaking (or even rigorously speaking, if one is allowed to use non standard analisys), a closed polygon with infinitely many edges of lenght $dx$. This is an example figure with $n=10$ edges: \$e^{ikx}...

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