# Tag Info

1

It's just replacing $|X|$ with $X$ or with $-X$ in ranges where $X$ is positive or negative, respectively, according to the modulus function definition.

2

I should also add that this integral may be evaluated using the Residue theorem by considering $$\oint_C dz \frac{\log^2{z}}{z^2+4}$$ where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$. In the limits as $R \to \infty$ and $\epsilon \to 0$, the contour integral is equal to $$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2+4} + ... 0 I believe that the problem assumes that you already know that$$\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}$$(or possibly a small variation of this) and your mission is to evaluate$$\int_{-\infty}^{\infty} e^{-x^2/3}dx$$by using a substitution to express a definite integral involving e^{-x^2/3} into one involving e^{-x^2}. Let's try. First, note ... 2$$A=\int_0^{\infty}\frac{\ln x}{x^2+4}dx\stackrel{x\to 2u}{=}\frac{1}{2}\left (\int_0^{\infty}\frac{\ln 2}{u^2+1}du+\int_0^{\infty}\frac{\ln u}{u^2+1}du\right )$$the first integral you can use u=\tan v the second one \int_0^{\infty}=\int_0^{1}+\int_1^{\infty}=I_1+I_2 for I_2=\int_1^{\infty} use v\to 1/v then you get I_2=-I_1 so our integral ... 2$$\begin{align}\frac{1}{2} \int\limits_0^\frac{\pi}{2} \ln (2\tan u) \,\mathrm{d}u & = \frac{1}{2} \int\limits_0^\frac{\pi}{2} \ln (2) \,\mathrm{d}u + \frac{1}{2} \int\limits_0^\frac{\pi}{2} \ln (\tan u) \,\mathrm{d}u\\~\\&= \dfrac{\pi \ln 2}{4} + \frac{1}{2} \int\limits_0^\frac{\pi}{2} \ln (\tan u) \end{align}$$Now notice below to conclude ... 1 Hint: If you had 1 instead of 4 in the denominator, what would the value be ? You could try a change of variable that's relevant both for \ln x and {\rm d}x/(1+x^2). 6 Because \ln x-1\geqslant 0 for x\geqslant \mathrm{e} and \ln x-1\leqslant 0 for x\leqslant \mathrm{e}. 2 Please check your question. The question you asked is different from the one shown in your reference. I am answering the question in your reference. \frac{\mathrm{d}}{\mathrm{d}x} \int_0^x (x-t)g(t)\,\mathrm{d}t=\frac{\mathrm{d}}{\mathrm{d}x} x\int_0^x g(t)-\int_0^x tg(t)\,\mathrm{d}t Then if you set F(x)=x\int_0^x g(t)， by chain rule, ... 2 Since (|f_n-f|)_{n\geqslant 1} is a sequence of non-negative functions we have$$ \int \sum_{n\geqslant 1} |f_n-f| \,\mathrm d\mu=\sum_{n\geqslant 1}\int |f_n-f|\,\mathrm d\mu<\infty. $$Therefore, \sum_{n\geq 1}|f_n-f|<\infty almost surely and hence also \lim_{n\to\infty}|f_n-f|\to 0 almost surely. 0 This question appears to have been answered in the comments, so I'm putting this to take it off the unanswered queue. Put y=x^2 \to dy =2xdx giving us \frac{1}{2}\int e^{-y^2}dy = \frac{\sqrt{\pi}}{4}\text{Erf}(y). --Winther 1 Think of it physically: each measure assigns different weights to given sets: consider for example the particular case d\mu=df(x)=f'(x)dx for a well behaved f(x). Here you can really see the difference between the "ordinary" measure dx, which does not care about the location of the set, and f'(x)dx, which indeed does! In formulas: ... 1 There are two inaccuracies. The first, which is innocuous, is that you have not specified if your integral starts at (1+) or (1-) (so that you catch, or not, a jump at the start), which could have led to a calculation error. The second is the calculation of the differential {\rm d}(\lceil x\rceil / x^{s+1}). There should be two parts : one coming from ... 6 Hint: Let u = \ln(x), then \frac{du}{dx} = \frac{1}{x} \implies du = \frac{dx}{x} If this is true, what do you see for$$\int (\ln(x))^{10}(\frac{dx}{x})$$? 0 Essentially we want to show:$$\int_a^bf(y)dy=\int_{g^{-1}(a)}^{g^{-1}(b)}f(g(x))g'(x)dx$$for a strictly increasing continuous function g (if g is decreasing, we take -g and absorb the minus sign into swapping the limits). If \{y_0,\ldots,y_n\} is a partition of [a,b], y_{j-1}\leq y_j^*\leq y_j, then a Riemann sum for the left integral is ... 0 Even if you would have meant to write e^{-x}, or \displaystyle\int_{-\infty}^0 , the integral would still not be expressible in terms of elementary functions and constants, since a simple substitution of the form kx^2=\sinh^2t would immediately create an expression in terms of Bessel and Struve functions, and their various derivatives. 0 yes, I think it was Leibniz (?) who introduced this ingenious "device" for making substitutions look nicer. When you make a substitution y=f(x) in an x-integral and you compute dy you really mean \displaystyle \frac{dy}{dx}\, dx but it looks nicer if you just "cancel" out the dx's even though that is not what happens. 0 Here is an alternative approach:$$\int_{-\infty}^\infty \frac{1}{1+x^2}=\oint_C \frac{1}{1+z^2}=2\pi i\mathrm{Res}(f(z),i)= 2\pi i\times \frac{-i}{2}=\pi$$where C is a counterclockwise contour. For your follow up question:$$\int_{-\infty}^{\infty}\frac1{\sqrt{1+x^3}}$$is not expressible in terms of elementary functions. 1 You were on the right track. We have$$\int_{-\infty}^0 \frac{1}{1 + x^2}\, dx = \lim_{a\to -\infty} \int_a^0 \frac{1}{1 + x^2}\, dx = \lim_{a\to -\infty} \arctan x\bigg|_{x = a}^0 = \lim_{a\to -\infty} (-\arctan a) = \frac{\pi}{2}$$and$$\int_0^\infty \frac{1}{1 + x^2}\, dx = \lim_{b \to \infty}\int_0^b \frac{1}{1 + x^2}\, dx = \lim_{b\to \infty} ...

1

You have $$\int_{-X}^{X}\frac1{(1+x^2)}dx=2\int_{0}^{X}\frac1{(1+x^2)}dx= 2 \arctan X$$ then use $$\lim_{X\to+\infty}\arctan X=\frac{\pi}{2}$$

0

$\iint_S f \operatorname d S$ is the Surface Integral of scalar field $f$ over surface $S$. When the curve can be described by a parameterised vector $S: \vec r(s,t)$, we have $$\iint_S f(\vec r(s,t))\operatorname d S = \iint_S f(\vec r(s,t))\begin{Vmatrix} \frac{\partial \vec r(s,t)}{\partial s}\times \frac{\partial \vec r(s,t)}{\partial t}\end{Vmatrix} ... -1 Use binomial expansions of the expressions in respective integrals as given in the previous solution of the problem. 1 \iint_A f(x,y)dA represents the volume underneath the surface z=f(x,y). When f(x,y)=1 you get the area of A, which is the volume under the plane z=1 intersected with A. Note that \iint_AdA represents a volume, and the value of this volume (the scalar part) is the area of A. 1 The result, with the indefinite integral, holds. It's almost the chain rule (the hypotheses that f is continuous isn't needed; so, the result seems strange). For definite integrals, the coresponding result isn't true. Take F(x)=x and g a differentiable function on [a,b] with g' non-integrable on [a,b] (see, again, the comments above). Then ... 2 Let f(x) = x^3. Given a partition P : a = x_0 < x_1 < \cdots < x_n = b of [a,b],$$x_{i - 1}^3 \le \frac{x_i^3 + x_i^2 x_{i-1} + x_ix_{i-1}^2 + x_{i-1}^3}{4} \le x_i^3 \quad (i = 1, 2, \ldots, n).$$Thus$$x_{i-1}^3(x_i - x_{i-1}) \le \frac{x_i^4 - x_{i-1}^4}{4} \le x_i^3(x_i - x_{i-1}) \quad (i = 1, 2, \ldots n).$$Taking the sum as i ... 0 Here it is an approach using partitions. We have:$$\int_{a}^{b}x^3\,dx = \int_{0}^{b}x^3\,dx -\int_{0}^{a}x^3\,dx,$$hence it is sufficient to prove that \int_{0}^{c}x^3\,dx = \frac{c^4}{4} or, by setting x=ct,$$\int_{0}^{1} t^3\,dt = \frac{1}{4}\tag{1}.$$Using Riemann sums over a uniform partition we have:$$\int_{0}^{1} t^3\,dt = \lim_{n\to ...

1

The form of the answer suggests integration by parts with the choice $$u = \cosh^{-1} \left( \frac{x}{2} + 1 \right), \quad dv = x^{-1/2} \, dx.$$ Then compute the derivative $$du = \ldots?$$ and the integral $$v = \ldots?$$ If you have trouble computing $du$, you can obtain it by writing $$x = \cosh u, \quad \frac{dx}{du} = \sinh u,$$ hence ...

3

By changing the variable $t=\frac\pi2-x$ the integral becomes $$\int_0^{\pi/2}\frac{t^\alpha}{(\sin t)^\beta}dt$$ and since $$\frac{t^\alpha}{(\sin t)^\beta}\sim_0\frac1{t^{\beta-\alpha}}$$ so the given integral is convergent if and only if $\beta-\alpha<1$.

0

$$\int \frac{(1+cot^{2} x) cot x}{csc(x)}= \int (1+cot^{2} x) cos x = -csc(x) + const$$

2

I suggest a different approach. First of all please note that $$sin(t)=\frac{e^{jt}-e^{-jt}}{2j}$$ and $$cos(t)=\frac{e^{jt}-e^{-jt}}{2}$$ Since $e^{jt}=z$ you can wtite your integral as $$-\frac{j(z-j)^2}{1-4 z+z^2}$$ while $dt= dz/(jz)$. Now your integral can be evaluates ...

1

Your contour integral is incorrect. It should be $$\int_{|z| = 1} \frac{1 - \frac{z - z^{-1}}{2i}}{2 - \frac{z + z^{-1}}{2}}\, \frac{dz}{iz}$$ which simplifies to $$-\int_{|z| = 1} \frac{z^2 - 2iz + 1}{z(z^2 - 4z + 1)}\, dz$$

2

In principle the integral is not too difficult using polar coordinates. I will do everything for $x_0=-x_1 ,y_0 =- y_1$ and $p=q=0$ to simplify notation a little bit. Then we don't have to shift and the whole area is given by $4$ times the integral over the first quadrant. We split the remaining integral in two regions. 1.) The triangle with vertices ...

0

Hint: $1 + \cot^2 x = \csc^2 x$. Simplifying may be enough to let you do it by inspection. If not, substitute $u = \sin x$ and $du = \cos x dx$.

1

If you don't want to appeal to Riemann-Lebesgue, you could also try integrating by parts (being careful to justify the differentiability at 0 of $u\mapsto (1-{\rm e}^{-u^4})/u^2$).

0

Hint: Check Riemann-Lebesgue lemma

1

$\cos x$ is negative on $[\pi/2,3\,\pi/2]$ and positive on $[3\,\pi/2,5\,\pi/2]$. $$\Bigl|\int_{\pi/2}^{3\pi/2}\frac{\cos x}{x}\,dx\Bigr|=\int_{\pi/2}^{3\pi/2}\frac{|\cos x|}{x}\,dx\ge\frac{2}{3\,\pi}\int_{\pi/2}^{3\pi/2}|\cos x|\,dx=\frac{4}{3\,\pi}\tag{1}$$ $$\int_{3\pi/2}^{5\pi/2}\frac{\cos x}{x}\,dx\le\frac{2}{3\,\pi}\int_{3\pi/2}^{5\pi/2}\cos ... 0 Try rewriting what's inside the radical. Maybe if you complete the square you can find something to substitute. 2 Write:$$\frac 1{\sqrt{4(x-1)^2 -1}}$$Can you see to substitute 2(x-1) = \sec\theta? Then 2\,dx = \tan\theta\sec\theta\,d\theta \iff dx = \frac 12 \tan\theta\sec\theta \,d\theta Recall here the identity$$\tan^2 \theta + 1 =\sec^2 \theta \iff \tan^2\theta = \sec^2\theta - 1$$Substituting, we get$$\begin{align}\int \frac 1{\sqrt{4(x-1)^2 -1}}\,dx ...

1

Hint: Note that $$4x^2-8x+3=4(x-1)^2-1$$

2

The thing you should think of instantly when you see a thing like that is that completing the square is the standard method in algebra for reducing a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial with no first-degree term. In this case $3-2x-x^2 = 4 - (x+1)^2 = 4-u^2$. Now you have a ...

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