3

According to Theorem 5.13 in Rudin's "Principles of mathematical analysis" (slightly restated) If $f$ and $h$ are real and differentiable in $(0,1)$ and if $h'(x)\neq 0$ in a neighborhood of $1$, and $h(x)\to +\infty$ as $x\to 1^-$, and in addition $f'(x)/h'(x)\to A$ as $x\to 1^-$, then $f(x)/h(x)\to A$ as $x\to 1^-$. Applying this to $$ f(x)=\...


3

HINT If you make the change of variable $y = \pi/2 - x$, you get \begin{align*} I = \int_{0}^{\pi/2}\frac{f(\cos(x))}{f(\cos(x)) + f(\sin(x))}\mathrm{d}x \end{align*} Then add both expressions of $I$ and see what happens.


3

Let $0<t<4$. With the substitution $u= \sqrt{x}$ we get $$\int _t^4\:\frac{e^{\sqrt{x}}}{2\sqrt{x}}dx= \int_{\sqrt{t}}^2 e^u du=e^2-e^{\sqrt{t}} \to e^2-e^0=e^2-1$$ for $t \to 0.$ Thus $$\int _0^4\:\frac{e^{\sqrt{x}}}{2\sqrt{x}}dx=e^2-1.$$


2

$$y=\frac{x^2}{1+x^2}=\frac{1+x^2-1}{1+x^2}=1-\frac{1}{1+x^2}=1-\frac{1}{(x+i)(x-i)}$$ $$y=1-\frac{i}{2 (x+i)}+\frac{i}{2 (x-i)}$$ Just finish


2

Forget explicit parameterization of $\gamma$, just use Stoke's theorem. In particular, use the version stated in complex coordinates. Let $E$ be the ellipse bounded by $\gamma$. Since $\gamma$ walks around $E$ in clockwise direction, it is "negative" to the orientation of $\partial E$, the boundary of ellipse. Apply Stoke's theorem in complex ...


2

$$\frac{dy}{dx}=\frac{(x+y)^2}{(x+2)(y-2)}$$ Let $x-2=X, y-2=Y$, then $$\frac{dY}{dX}=\frac{(X+Y)^2}{XY}$$ Let $$Y=VX \implies \frac{dY}{dx}=X\frac{dV}{dX}+V$$ $$\implies X\frac{dV}{dX}+V= \frac{(1+V)^2}{V} \implies X\frac{dV}{dX}=\frac{(1+V)^2}{V}-V \implies \int \frac{VdV} {1+2V}= \int \frac{dX}{X}$$ $$\implies V=2\ln[CX(1+2V)^{1/4}], V=Y/X, X=x+2. Y=y-...


1

With usual shperical coordinates you would get it with $\sin(\varphi)$, so $$\int_0^{\pi} \int_{0}^{2\pi}\int_{t}^{1}(r^2)^{-\frac{p}{2}} \cdot r^2 \sin(\varphi) \ dr \ d\theta \ d\varphi=$$ $$= 2\pi \biggl[ -\cos(\varphi) \biggr]_0^{\pi}\int_{t}^{1}r^{2-p} \ dr =$$ which converges if and only if $p-2<1$ which means $p<3$, an it is: $$= 4\pi\int_{t}^{1}...


1

$$ z(t) = 2 \cos t + i3 \sin t$$ Explanation: $$ z(t) = x(t) + i y(t)$$ Now, for the ellipse for a trignomeetric parameterization: $x=2 \cos t$ $y=3 \sin t$, plug that same thing into the complex function in $t$


1

Using the generating funtion $e^{2xt-t^2}=\sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}$ we can obtain two recurrence relations. Differentiating with respect to $t$ we obtain $H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x)$ for $n\geq 1$ with $H_0(x)=1$ and $H_1(x)=2x$. Similarly, differentiating with respect to $x$ we obtain $\frac{d}{dx}H_n(x)=2nH_{n-1}(x)$ for $n\geq 1$. Now, ...


1

It seems to me that the assumption on $K$ should be that $$\int^\infty_0 |K(y)|\,y^{-1/p}\,dy<\infty$$ As you pointed out, the generalized Minkowski inequality gives \begin{align} \left(\int^\infty_0\left|\int^\infty_0 K(y) f(xy)\,dy\right|^p\,dx\right)^{1/p} &\leq \left(\int^\infty_0\left(\int^\infty_0 |K(y)| |f(xy)|\,dy\right)^p\,dx\right)^{1/p}\\ &...


1

It's your second one that is wrong: you need to integrate the region under $y=1+x^3$ so you'd be best served using the shells formula in that case. But as you mentioned in the comments you can't use that formula. That's okay, but now you'd need two integrals to get the answer. First do the lower portion which is just the region $1 \leq x \leq 2$ and $0 \...


1

By the reverse triangle inequality $|\,|f(u) - |f(v)|\,| \leqslant |f(u) - f(v)|$. Hence, for any partition interval $I_j = [x_{j-1},x_j]$ we have $$M_j(|f|,P)- m_j(|f|,P) = \sup_{u,v \in I_j}|\,|f(u)| - |f(v)|\,| \leqslant \sup_{u,v \in I_j}|f(u) - f(v)| = M_j(f,P)- m_j(f,P),$$ and $$U(|f|,P) - L(|f|,P) = \sum_{j=1}^n(M_j(|f|,P)- m_j(|f|,P)) \Delta x_j \...


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