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14

$$L=\lim_{x\to0}\frac{x-\sin x}{x^3}\\ =\lim_{x\to0}\frac{2x-\sin2x}{8x^3}\\ 4L-L = \lim_{x\to0}\frac{x-\frac12\sin2x-x+\sin x}{x^3}$$ which simplifies to the product of three expressions of the form $\frac{\sin y}{y}$


7

Let $\displaystyle \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathcal I=\int_0^\infty\frac{1}{1+x^6} \,\mathrm dx $ $$\begin{align} I&=\frac{1}{2}\left[ \int_0^\infty \frac{(1-x^2+x^4)+x^2+(1-x^4)}{(1+x^2)(1-x^2+x^4)} \,\mathrm dx \right]\tag{1}\\ &= \frac{1}{2}\left[\int_0^\infty \frac{1}{1+x^2} \,\mathrm dx + \int_0^\infty ...


6

To answer your questions (1) Integrating along a contour means integrating along a parametrized path, i.e. given a parametrization $\gamma: [0,1]\to \Bbb C$ which has the property that $\gamma([0,1])=C$, the contour, integrating along the path means to compute $$\int_C f=\int_0^1(f\circ\gamma)(t)\gamma'(t)\,dt$$ (2) $z$ is the argument of the function, ...


3

Recall that $\mathcal B(\Bbb R)$ has the same cardinality as $\Bbb R$, namely $2^{\aleph_0}$. Since $M$ is a set of functions between the two sets, its cardinality is bounded by the cardinality of $\Bbb{R^R}$ which is $2^{2^{\aleph_0}}$. On the other hand, given any subset $A\subseteq \Bbb R$, we can consider $\mu_A$ which is an atomic measure satisfying ...


3

how can this be done WITHOUT complex analysis? $\quad$ All integrals of the form $~\displaystyle\int_0^\infty\frac{x^{k-1}}{(x^n+a^n)^m}dx~$ can be evaluated by substituting $x=at$ and $u=\dfrac1{t^n+1}$ , then recognizing the expression of the beta function in the new integral, and lastly employing Euler's reflection formula for the $\Gamma$ function ...


2

It follows from your equation (1) that $f$ is differentiable and $f(x) = \lambda f'(x)$ for all $x \in \mathbb R$. The general solution of this differential equation is $f(x) = C \exp(x/\lambda)$ for some $C \in \mathbb R$. Since $f(0) = 0$, $f$ is identically zero.


2

$\sin x = x - \dfrac{x^3}{3!} + o(x^5) \to \dfrac{x-\sin x}{x^3} = \dfrac{1}{6} + o(x^2)$. From this the limit is $\dfrac{1}{6}$.


2

I suppose you can always use Taylor expansion near $x=0$ (as long as it is known to you), that is: $$ \sin x = x - \frac1{3!}x^3 + o(x^5) $$ Hence we will have: $$ \lim\limits_{x\to 0}\frac{x-\sin x }{x^3}=\frac16 $$


2

Since the function $$ x \longmapsto\frac{1}{e^x-1} $$ is continuous on $(0,1]$, a potential convergence problem, concerning the integral, is for $x$ near $0$. We have $$ \frac{1}{e^x-1} \sim_{0^+} \frac1x $$ thus your integral $\displaystyle \int_0^1 \frac{1}{e^x-1} dx$ is divergent as $\displaystyle \int_0^1 \frac{1}{x} dx$ is divergent. In the same ...


1

If $X = [0,T]$ and $\mu$ is the Lebesgue measure on $X$, then by Cauchy-Schwarz $$\|f\|_1 \le \mu(X)^{1/2} \|f\|_2 = \sqrt{T}\|f\|_2$$ (Note: Your inequality states $\|f\|_1 \le \mu(X)^{-1/2}\|f\|_2$, which is not always true. For example, if $f(t) = 1$ and $T = 2$, then $\|f\|_1 = 2$ and $\mu(X)^{-1/2}\|f\|_2 = 1 < 2 = \|f\|_1$.)


1

As Daniel Fischer pointed out in the comments, the claim is false and a counter example is $f(x) = x^3\sin(\frac{1}{x})$.


1

Because $\mathbb{R}$ is complete separable, all Borel measures are regular so they are determined by their values on finite unions of rational intervals. It follows that there are continuum many Borel measures on reals.


1

$\cos z = \Re(e^{iz})$ and $\Re$ is a linear operator, so $\Re\int = \int\Re$.



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