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## Hot answers tagged integration

6

$$\frac{\pi^2}{24}-\frac{2\pi}3+\frac1{36\sqrt{2}}\left[5\pi^2+12\left(4+\ln\left(\frac{1+\sqrt2}2\right)\right)\left(\pi-2\ln\left(1+\sqrt2\right)\right)-48\operatorname{Li}_2\left(\sqrt2-1\right)\right]$$

5

Often, with such complicated primitives, I suggest not to calculate the primitive, but to use some trick, like introducing a parameter and differentiate with respect to it. Read below if you are interested in such a way to calculate this integral. Tell me if this is far from what you looked for. First, since the integrand is even, your integral equals $$... 5 Hint. Observe that we have$$ \begin{align} \int_{-\infty}^\infty \frac{x^2 e^x}{(e^x+1)^2}\:dx&=\int_{-\infty}^0 \frac{x^2 e^x}{(e^x+1)^2}\:dx+\int_0^\infty \frac{x^2 e^x}{(e^x+1)^2}\:dx\\\\ &=2\int_0^\infty \frac{x^2 e^x}{(e^x+1)^2}\:dx\\\\ &=2\int_0^\infty \frac{x^2e^{-x} }{(1+e^{-x})^2}\:dx\\\\ ...

4

Consider $h>0$ then there is $n$ so that $\frac{1}{(n+1)\pi +\pi/2} < h\le \frac{1}{n\pi + \pi/2}$. In this interval (called $I_n$), $\cos(1/t)$ is positive (resp. negative) if $n$ is odd (resp. even). Also \int_0^h \cos\left(\frac 1t\right) dt = \sum_{k=n+1}^\infty \int_{I_k} \cos\left(\frac 1t\right) dt + \int_{\frac{1}{(n+1)\pi + \pi/2}}^h ... 3 Partial integration \begin{align} \int \cos \bigg(\frac{1}{t} \bigg) dt &= \int t^2 \frac{1}{t^2}\cos \bigg(\frac{1}{t} \bigg)dt \\ &= -t^2 \sin \bigg(\frac{1}{t} \bigg) + \int 2 t \sin \bigg(\frac{1}{t} \bigg) dt \\ \end{align} 3 It looks like a quadratic because it is a quadratic: \begin{align} 0\le\int_0^1(f-\lambda x)^2&=\int_0^1f^2-2\lambda\int_0^1xf+\lambda^2\int_0^1x^2\\ &=\int_0^1f^2-2\lambda\int_0^1xf+\lambda^2\cdot \frac{1}{3}. \end{align} So the minimum is 0 and it is attained when f(x)=\lambda x. 3 Following my comment above: the function f(x,a) = \cos(a x) is regular enough (continuously differentiable, at least) for your approach of switching integral and differentiation to be justified. The error is in the last step: g\colon a\mapsto\int_0^{2\pi} f(x,a)dx = \int_0^{2\pi} \cos ax dx is a function of a, which is not identically zero. Computing ... 3 If the Fourier Transform is interpreted in the Cauchy Principal Value sense, then we have\begin{align} f(t)&=\frac1{2\pi}\int_{-\infty}^\infty \frac{e^{i\omega t}}{\tau \omega}\,d\omega\\\\ &=\frac1{2\pi}\lim_{L\to \infty,\epsilon\to 0^+}\left(\int_{-L}^{-\epsilon} \frac{e^{i\omega t}}{\tau \omega}\,d\omega+\int_{\epsilon}^{L} \frac{e^{i\omega ...

2

Just for some new ideas! I would reccomend a completely different method. This method uses the Gudermannian $\text{gd}$ function. So you would substitute $x=\text{gd}(a);\text{d}x=\text{sech}\space a\text{d}a$ That transforms the integral into: $$\int \frac{\tanh a}{\tanh a+\text{sech}\space a}(\text{sech}\space a)\mathrm da$$ Through some hyperbolic trig ...

2

For the first: complete the square to rewrite this as $$\sqrt{(x+3/2)^2 + 3/4}$$ Now, apply a substitution $\theta$ such that $$x+3/2= \sqrt{3/4}\tan \theta$$ For the second: complete the square to rewrite the integrand as $$\sqrt{k - (x-3/2)^2}$$ For some $k$ that I'm too lazy to calculate. Then, apply a substitution such that $$x-3/2= \sqrt k \sin ... 2 By a linear transform of the argument, you can normalize the integrand to one of$$\sqrt{1+x^2}\text{ or }\sqrt{1-x^2}.$$Then by parts,$$\int\sqrt{1\pm x^2}\,dx=x\sqrt{1\pm x^2}\mp\int\frac{x^2}{\sqrt{1\pm x^2}}dx=x\sqrt{1\pm x^2}-\int\frac{(1\pm x^2)-1}{\sqrt{1\pm x^2}}dx.$$Move the first part of the integral to the LHS and remains to integrate ... 2 Take g=1_{[0,1]^d}, f= \sum_{n=0}^\infty n \cdot 1_{\{n\}}, then f is unbounded, but \int f = 0 and \int g = 1. 2 Let g=\frac1h and s=\frac1t. Then integration by parts gives$$ \begin{align} \lim_{h\to0}\frac1h\int_0^h{\cos\!\left(\frac1t\right)\mathrm{d}t} &=\lim_{g\to\infty}g\int_g^\infty{\frac{\cos(s)}{s^2}\,\mathrm{d}s}\\ &=\lim_{g\to\infty}g\left[-\frac{\sin(g)}{g^2}+2\int_g^\infty\frac{\sin(s)}{s^3}\,\mathrm{d}s\right]\\ ...

2

HINT: $$\int\frac{1}{\left(t+\cos(a)\right)^2+\sin^2(a)}\space\text{d}t=$$ Substitute $u=\cos(a)+t$ and $\text{d}u=\text{d}t$: $$\int\frac{1}{\sin^2(a)+u^2}\space\text{d}u=$$ $$\int\frac{\csc^2(a)}{u^2\csc^2(a)+1}\space\text{d}u=$$ $$\csc^2(a)\int\frac{1}{u^2\csc^2(a)+1}\space\text{d}u=$$ Substitute $s=u\csc(a)$ and ...

1

Factoring gives $$\frac{1}{\sin^2 \alpha} \int\frac{dt}{(t \csc \alpha + \cot \alpha)^2 + 1},$$ so that the denominator of the integral has the familiar form $v^2 + 1$. This suggests writing $$t \csc \alpha + \cot \alpha = v = \tan u,$$ and multiplying both sides of this substitution by $\sin \alpha$ gives the desired substitution: ...

1

You cannot have $$\int x^5+6\,dx = \int x^5dx + 6.$$ What you can have is $$\int x^5+6\,dx = \int x^5\, dx +\int 6\,dx.$$ This is am important property know regarding integrals. Addendum: You can also have $$\int x^5+6\,dx = 6\int\frac{1}{6}x^5+1\,dx.$$ These a basic properties of integrals. Try reviewing them. This link might be helpful.

1

For any partition $\mathcal{P} = \{a = t_0 < t_1 < \cdots < t_n = b\}$ of $[a,b]$ we know that $$\int_a^b f \le U(f, \mathcal{P})$$ where $$U(f, \mathcal{P}) = \sum_{i=0}^n \sup_{[t_{i-1}, t_i]} f(x) (t_i - t_{i-1})$$ is the upper Riemann sum of $f$ with the partition $\mathcal{P}$. Furthermore, as you state in your question, you know that $$... 1 Using Feynman trick plus the Real part trick:$$\Re\lim_{\alpha \to 1}\int_{0}^{2\pi} \frac{\partial^4}{\partial \alpha^4} e^{i\alpha x}\text{d}x = \Re\lim_{\alpha \to 1}\frac{\partial^4}{\partial \alpha^4}\left(\frac{e^{2\pi i\alpha} - 1}{i \alpha}\right) $$Calculate the derivative, obtaining:$$\Re\lim_{\alpha \to 1} \left(\frac{8\left(3i + e^{2\pi i ...

1

The formula for the moment of inertia about the $z$-axis is $$I_z=\int_V\rho(r) z^2\,\mathrm{d}V$$ Here, $\rho$ is density, $r$ is radius, and $V$ is volume. You're given that $$\rho(r)=1-r^2$$ Let's do some coordinate changes. In spherical coordinates, $$z=r\cos\theta$$ The volume element is ...

1

Let $\alpha=\arctan\sqrt{\frac{x}{2}}$ $$I=\int \frac{\alpha dx}{\sqrt{x+2}}=\int\alpha d(2\sqrt{x+2})=2\alpha\sqrt{x+2}-2\int \sqrt{x+2}\space d\alpha$$ The calculation gives $$d\alpha=\frac{\sqrt {2} dx}{\sqrt x(x+2)}$$ Hence $I=2\alpha\sqrt{x+2}-2\sqrt 2\int \frac{\sqrt{x+2}\space dx}{(x+2)\sqrt x}$ $I=2\alpha\sqrt{x+2}-2\sqrt ... 1 For the$y<0$case your DE is evaluated incorrectly: $$\frac{dy}{dx}=\sqrt{-y}$$ $$(-y)^{-\frac{1}{2}}\frac{dy}{dx}=1$$ $$-2\sqrt{-y}=x+c$$ $$-4y=(x+c)^2$$ $$y=-\frac{(x+c)^2}{4}$$ This should hopefully help you to visualize the direction field now. Something like this: 1 Actually, the Fundamental Theorem of Calculus states that if$f:[a,b]\to \mathbb R$is a continuous real-valued function, and$F$is its antiderivative, then $$\int_a^x f(t)dt= F(x)-F(a)$$ This is precisely why, more accurately, $$\int_a^x f(t)dt= F(x)-F(a)=\int f(x)dx +C$$ for some$C \in \mathbb R$. 1 You are probably referring to this part of the fundemental theorem of calculus $$\frac{d}{dx} \int_a^x f(t)dt =f(x)$$ And your using the fact that $$f(x) = \frac{d}{dx} \int f(x) dx$$ which yields $$\frac{d}{dx} \int_a^x f(t)dt = \frac{d}{dx}\int f(x) dx$$ While this is true, ... 1 For any integral of the forms, you have mentioned, transform the integrand into the sum or difference of two perfect square expressions and the resulting expression will look like$\sqrt{(x\pm a)^2+(b)^2}$,$\sqrt{(x\pm a)^2-(b)^2}$or$\sqrt{(b)^2-(x\pm a)^2}$. And then use the following formulae which can be deduced with the help of integration by ... 1 The integrand has a closed-form antiderivative in terms of elementary functions and dilogarithms. Mathematica can find it if we help it by first converting the arctangent into a combination of logarithms: $$\arctan(x)=\frac i2\ln(1-i x)-\frac i2\ln(1+ix)$$ After some simplifications it takes this form. Its correctness can be checked manually using direct ... 1 HINT: Using eliptic integrals! $$\int\sqrt{1+\cos^2(2x)}\space\text{d}x=\frac{\text{E}\left(2x\mid\frac{1}{2}\right)}{\sqrt{2}}+\text{C}$$ 1$f(x) = \frac{\sin^2(x)}{\tan(x)}=\sin(x)\cos(x)=\frac{\sin(2x)}{2}f'(x) = \cos2x$Arc length,$l=\int\sqrt{1+(f'(x))^2} dx = \int\sqrt{1+\cos^2(2x)} dx$This is a non elementary integral. 1 The idea here is that all but finitely many of the points in$\{ 1/n : n \in \mathbb{N} \}$, where the discontinuities in$f$are, are very close to$0$. So you can work this way. Let$\varepsilon >0$. Choose$N \in \mathbb{N}$such that$1/(N+1)<\varepsilon/2$. Then make the first two points of the partition be$0$and$1/(N+1)$. Now you just have ... 1 Because$\frac{x^2e^x}{(e^x+1)^2}=\frac{x^2}{\left(e^{x/2}+e^{-x/2}\right)^2}$is even, we can integrate by parts and use the Dirichlet eta function:$\$ \begin{align} \int_{-\infty}^\infty\frac{x^2e^x}{(e^x+1)^2}\,\mathrm{d}x &=2\int_0^\infty\frac{x^2e^x}{(e^x+1)^2}\,\mathrm{d}x\\ &=-2\int_0^\infty x^2\,\mathrm{d}\frac1{e^x+1}\\ ...

1

Most likely the problem that you have posed will not help at all in the situation you describe, but it is an interesting problem with interesting answers and so, if you approach this as a side topic that offers you a holiday from your real work you might be entertained. First the problem of when a function is a derivative (pointwise everywhere) is an old ...

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