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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: \begin{equation} ...


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

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


2

I suggest a different approach. First of all please note that \begin{equation} sin(t)=\frac{e^{jt}-e^{-jt}}{2j} \end{equation} and \begin{equation} cos(t)=\frac{e^{jt}-e^{-jt}}{2} \end{equation} Since $e^{jt}=z$ you can wtite your integral as \begin{equation} -\frac{j(z-j)^2}{1-4 z+z^2} \end{equation} 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 ...


1

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1

Since this post was tagged with complex analysis, I can provide a contour integration solution as well. Again, I will exploit the identity given to you by @idm. $$ \int_0^{\pi/2}\ln(\sin(\theta))d\theta = \frac{1}{2}\int_0^{\pi}\ln(\sin(\theta))d\theta $$ Consider $1 - e^{2iz} = -2ie^{iz}\sin(z)$. We can write $1 - e^{2iz}$ as $$ 1 - e^{-2y}(\cos(2x) + ...


1

As shown by @idm, we have that $$ \int_0^{\pi/2}\ln(\cos(x))dx = \int_0^{\pi/2}\ln(\sin(x))dx. $$ We can exploit this identity for another one and use Feynman's method (differentiating under the integral). Consider $$ \int_0^{\pi/2}x\cot(x)dx.\tag{1} $$ By integration by parts, we have $$ \int_0^{\pi/2}x\cot(x)dx = x\ln(\sin(x))\Bigr|_0^{\pi/2} - ...


5

HINT: Complete the square $$3 - 2x - x^2 = 4-1-2x-x^2 =2^2 - (x+1)^2$$ Then put $(x+1) = 2\sin \theta$.


0

From central section containg unit sphere centers, $ r^2 = 1^2 -(x/2)^2, r= \sqrt{1- x^2/4}$ , circumference = $ 2 \pi r $


-1

It is a perfect answer to the question where it exploits the given hint First write the integral as (since the integrand is an even function) $$ I = \int_{-\infty}^\infty e^{-\frac{x^2}{3}}\,\mathrm dx = 2\int_{0}^\infty e^{-\frac{x^2}{3}}\,\mathrm dx $$ then exploit the hint you have been given which is the change of variable $u=x^2/3$ then use ...


5

Let $$I=\int_{-\infty}^{\infty}\exp(-\frac{x^2}{3})dx$$. $$I^2=\int_{-\infty}^{\infty}\exp(-\frac{x^2}{3})dx\int_{-\infty}^{\infty}\exp(-\frac{y^2}{3})dy$$ $$=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-\frac{x^2+y^2}{3})dxdy$$ $$=\int_{0}^{2\pi}\int_{0}^{\infty}\exp(-\frac{r^2}{3})rdrd\theta$$ ...


2

hint: what's the density of a standard (normal) distribution? take $y := cx$ with appropriately chosen $c$ (and use the fact that integral of a density is $1$)


1

There's no need to use an integral. Looking at the cross section gives a right triangle whose hypotenuse is $1$ (distance from the center of a sphere to a point of intersection), whose base is $\frac{x}{2}$ (distance from center of sphere to center of circle of intersection), and hence the other leg in this triangle is $\sqrt{1 - \frac{x^2}{4}}$ and this is ...


0

Form a system with the implicit equations of the spheres $$x^2+y^2+z^2=r_0^2,\\(x-d)^2+y^2+z^2=r_1^2.$$ (Note that I used $d$ instead of your $x$.) Subtracting an equation from the other, the other turns to the plane of support of the circle. $$2dx-d^2=r_0^2-r_1^2,$$ or $$x=\frac{r_0^2-r_1^2+d^2}{2d}.$$ and this gives you the equation of the circle, ...


0

You may like this method. Note $$ \lim_{n\to\infty} n(x^{\frac{1}{n}}-1)=\ln x, \text{ for }x>0 $$ and hence, for $m\in\mathbb{N}$, \begin{eqnarray} \int_0^1\frac{x^m-1}{\ln x}dx&=&\int_0^1\sum_{i=0}^{m-1}\frac{x^i(x-1)}{\ln x}dx\\ &=&\int_0^1\sum_{i=0}^{m-1}\lim_{n\to\infty}\frac{x^i(x-1)}{n(x^{\frac{1}{n}}-1)}dx\\ ...



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