19

This is not necessarily true. Take the following example; $$\int_a^{2a}\frac1x\mathrm{d}x=[\ln{|x|}]_a^{2a}=\ln{(2)}$$ If we take $a\to\infty$ then the integral becomes $$\int_\infty^\infty\frac1x\mathrm{d}x=\ln{(2)}$$ as the integral is constant for all $a\in\mathbb{R}$. What I guess your professor meant was that $$\lim_{a\to\infty}\int_a^a f(x)\mathrm{d}x=...


7

Really WolframAlpha is just wrong here. An antiderivative must be absolutely continuous, and $F(x) = -\cos(x)\mathrm{sgn}[\sin(x)]$ isn't. The correct antiderivative (up to an additive constant) is $$ F(x) = 2\left\lfloor\frac{x}{\pi}\right\rfloor - \cos(x)\mathrm{sgn}[\sin(x)] $$ which is absolutely continuous and satisfies $\int_a^b |\sin(x)|dx = F(b) - ...


6

An improper integral with an endpoint of $\infty$ means a limit of proper integrals where the endpoint approaches $\infty$. Thus a reasonable definition of $\int_{\infty}^\infty f(x)\; dx$ would be $$ \int_{\infty}^\infty f(x)\; dx = \lim_{a, b \to \infty} \int_a^b f(x)\; dx $$ This is $0$ if and only if $\int_a^\infty f(x)\; dx$ converges for some $a$. ...


6

$$\left| \int_{-2\pi}^{2\pi}x^2\sin^8(e^x)dx\right|\leq \int_{-2\pi}^{2\pi}\big|x^2\sin^8(e^x)\big|dx\le\int_{-2\pi}^{2\pi}x^2dx=2\int_0^{2\pi}x^2dx=2\cdot \frac{(2\pi)^3}{3}=\frac{16\pi^3}{3}$$ since $\big|\sin^8(e^x)\big|\le 1$ for all $x$, in particular, for $-2\pi\leq x\leq 2\pi$.


4

By the mean-value property of harmonic functions, $f(z)$ is the average of $f$ over any circle centred at $z$, and therefore over any disk centred at $z$. A disk $D$ of radius $r$ centred at $z$ has area $\pi r^2$, so $$\int_D |f(x+iy)|\; dx\; dy \ge |f(z)| \pi r^2$$ and as $r \to \infty$ the only way to avoid an infinite limit is $f(z)=0$.


4

One way is using the expansion of $\operatorname{csch}x$, that is $$\operatorname{csch}x=\dfrac{1}{x}+\sum_{n=1}^{\infty}\dfrac{2(-1)^nx}{n^2\pi^2+x^2}$$ then $$\int_{0}^{\infty}\dfrac{\operatorname{csch}x-\frac1x}{x}\ dx=\sum_{n=1}^{\infty}\int_{0}^{\infty}2(-1)^n\dfrac{1}{n^2\pi^2+x^2}\ dx=\sum_{n=1}^{\infty}\dfrac{2(-1)^n}{n\pi}\arctan\dfrac{x}{n\pi}\Big|...


4

$$I=\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a+b\tan(x))\mathrm dx\overset{x\to -x}=\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos(a-b\tan(x))\mathrm dx$$ Summing up the two integrals from above gives us: $$ 2I= 2\cos a \int_{-\pi/2}^{\pi/2} \cos^2 x \cos(b\tan x) dx\Rightarrow I=2\cos a\int_0^{\pi/2}\cos^2 x\cos(b\tan x) dx$$ $$\overset{\tan x=t}=2\cos a\int_0^\infty \frac{\...


3

Define the function $F$ for $x>0$ by: \begin{align}F(x)=\text{cotanh}\left(\frac{x}{2}\right)-\frac{2}{x}\end{align} Observe that, \begin{align}\lim_{x\rightarrow 0} F(x)&=0\\ \lim_{x\rightarrow \infty} F(x)&=1\\ F(x)-F(2x)&=\frac{1}{\sinh x}-\frac{1}{x} \end{align} On can use Frullani's theorem: \begin{align}\int_0^\infty \frac{\text{csch}(...


4

Hint: Write your integral in the form $$\frac{1}{x}\int\frac{1}{1+\left(\frac{y}{x}\right)^2}dy$$ and Substitute $$t=\frac{y}{x}$$ The result is given by $$\arctan \left( {\frac {y}{x}} \right) +C$$


3

The Jacobian is $1$, not $(-1)^{n}$. (You forgot the absolute value). So $\int f(x-y)g(y)dy=\int f(y)g(x-y)dy$ and $(f*g)(x)=(g*f)(x)$.


3

From the change of variable $u=x^{-p}$, the integral becomes $$ -\frac{1}{p}\int_{0}^\infty \sin(u)u^{1/p-1}du=-\frac{1}{p}\mathcal{M}\{\sin(u)\}(1/p), $$ where $\mathcal{M}$ denotes the Mellin transform. Since $0<1/p<1$, one can easily infer from http://mathworld.wolfram.com/MellinTransform.html that $$ -\frac{1}{p}\Gamma\left(\frac{1}{p}\right)\sin\...


3

Here's a sketch of how one might do this: fix $\epsilon>0$ and take a given $x$. To show continuity, we must find a $\delta>0$ such that $\vert g(y)-g(x)\vert<\epsilon$ when $\vert y-x\vert\leq \delta$. First, use the polynomial growth condition on $f$ to show that there exists some constant $K$ (depending only on $x$ and $\epsilon$) such that for ...


3

Note that $$U_n = \left\{ f\in L^2([0,1]) : \int_0^1 f f_n dx\neq 0 \right\}$$ is an nonempty (since $f_n$ is nonzero) open sets which is dense in $L^2 ([0,1])$. The Baire Category theorem says that $$ \bigcap U_n$$ is nonempty. Thus there is $g\in L^2([0,1])$ so that $$ \int_0^1 g f_n dx\neq 0$$ for all $n$.


3

As has been pointed out by other answers, this is not always true because the symbol $\infty$ can hide many things, even if we work with the extended reals. The actual meaning of the $\infty$ is a limiting process as a certain variable becomes arbitrarily large. The upper and lower limits in the $$\int_a^b f(x)\mathrm d x$$ however may approach $\infty$ at ...


2

The confusion seems to stem from the fact that the same symbol is used for the constant of integration in both integrals. If we instead write the general antiderivatives using the (a priori distinct) arbitrary constants $C, D$, our equality is then $$\operatorname{arccot} x + C = -\arctan x + D ,$$ and rearranging gives that $$\arctan x + \operatorname{...


2

The comment above is spot on, use $$\sin(x)\leq 1$$ So it must be that $$\int_{-2\pi}^{2\pi} x^2\sin^8(e^x)dx\leq\int_{-2\pi}^{2\pi} x^2dx $$ This has reduced the integration to a much simpler form. Proceed from here to find your answer. In questions like this, one generally wants to find the maximum of the integrand, or the maximum of one piece of ...


1

Setting $I_{a,b} $ equal to the original integral and assuming $b > 0$, we get, with your substitution $b\tan(x) = u-a$, $$\begin{split} I_{a,b} &= \frac 1b \int_{-\infty}^{+\infty} \cos^4(x) \cos(u) \mathop{}\!du = \frac 1b \int_{-\infty}^{+\infty} \frac{\cos(u) \mathop{}\! du}{\sec^4(x)} = \frac1 b\int_{-\infty}^{+\infty} \frac{\cos(u)\mathop{}\! du}...


1

Write $$2x=u,\quad y-x=v,\qquad {\rm resp.,}\qquad x={u\over2},\quad y={u\over2}+v\ .$$ Then $D$ is the linear image of $\hat D=\bigl\{(u,v)\bigm| u^2+v^2\leq R^2\bigr\}$. Since the Jacobian $x_uy_v-x_vy_u\equiv{1\over2}$ we have $${\rm area}(D)={1\over2}{\rm area}(\hat D)={\pi\over2}R^2\ .$$


1

Write the equation of the curve as $$y^2-2xy+5x^2-R^2=0$$ this gives $$y=x\pm\sqrt{R^2-4x^2}.$$ So the required area is area between $y=x\pm \sqrt{R^2-4x^2}.$ from $x=-R/2$ to $x=R/2$.We get the required area $A$ by $$A=2\int_{=R/2}^{R/2} \sqrt{R^2-4x^2} dx==2\int_{0}^{R} \sqrt{R^2-z^2} dz=\pi R^2/2.$$ The given curve is an ellipse.


1

In my experience, changing the order of integration when the limits are variable is sometimes possible, but never intuitive. Plotting the region is one of those kinds of steps that I can't skip without getting something wrong, so I simply don't skip that step. Ever. It is, admittedly, even harder in 3D than in 2D, but I would still recommend plotting the ...


2

To calculate centroid of a curve, first we compute the $\mathrm{d}s$: $$\mathrm{d}s = \sqrt{x^\prime(t)^2+y^\prime(t)^2+z^\prime(t)^2}=\sqrt{e^{2t}+2+ e^{-2t}}$$ Now note that $$\int_0^1 \mathrm{d}s = \int_0^1 \sqrt{e^{2t}+2+ e^{-2t}} \mathrm{d}t = \int_0^1 \sqrt{e^{-2t}(e^{4t}+2e^{2t}+1)} \mathrm{d}t = \int_0^1 e^{-t}(e^{2t}+1)\mathrm{d}t$$ $$=\int_0^1 e^...


1

Do you consider $L^p(X,\mathcal{A},\mu)$ to be real or complex valued functions? Let us consider real functions. Then $g\in L^\infty$ just corresponds to $(g(0),g(1)) \in \mathbb{R}^2$ which you could identify with $\mathbb{C}$. Hence you can identify the dual space of $L^\infty$ with $\mathbb{R}^2$ (or $\mathbb{C}^2$) itself. Let $f\in L^1$ and $g\in L^\...


1

$$l=R\tan\alpha\implies dl=\frac{R\,d\alpha}{\cos^2\alpha}.$$


1

If $f$ is $C^1$ and supported on $[-A,A]$ for $T >2A, t \in [-T/2,T/2]$ we have the Fourier series $$ f(t) = \frac{1}{T}\sum_n c_n(T) e^{2i \pi n t/T}, \qquad c_n(T) = \int_{-T/2}^{T/2} f(t)e^{-2i \pi nt/T}dt$$ Let $$F(u) = \int_{-\infty}^\infty f(t) e^{-2i \pi tu}dt, \qquad c_n(T) = F(n/T)$$ If also $|F(u)| < c|u|^{-2}$ we have $$\int_{-\infty}^\...


1

I think your second confusion is only due to notation: Sometimes taking the integral of a function with respect to $x$ is written as $\int dxf(x)$ instead of $\int f(x) dx$. The idea behind this notation is that $\int dx$ represents some operator which is applied to whatever comes right to it. So this would mean that (1) is equal to $\frac{1}{2\pi}\int_{-\...


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