Challenge Math Challenge - October 2018

[Periwinkle]

8. Let $f$, $g$: $\mathbb{R} \rightarrow \mathbb{R}$ be two functions with $f\,''(x) + f\,'(x)g(x) - f(x) = 0$. Show that if $f(a) = f(b) = 0$ then $f(x) = 0$ for all $x\in [a,b]$. (by @QuantumQuest )

According to Weierstrass's maximum value theorem, if a function is continuous on $[a,b]$, then it is bounded on this interval and takes its maximum and minimum. If function $f(x)$ is not constant, then its maximum or minimum is different from $f(a) = f(b)=0$. Let this be the maximum that the function takes at an internal $x_0$ point of the interval. According to one of Fermat's theorems, in this case - provided that the function $f (x)$ can be derived at this point - the derivative $f'(x_0)$ here is $0$. Therefore, the following equation is correct:
$$f\,''(x_0) + 0 \cdot g(x_0) - f(x_0) = 0$$ $f(x_0)$ is positive, so $f\,''(x_0)$ is also positive. Therefore, the function $f(x)$ has a local minimum at $x_0$, which cannot be a local maximum.

 

[benorin]

4) b) $$\int_A \frac{1}{x^2+y}\, d\lambda(x,y) =\int_0^1\int_0^1 \frac{dydx}{x^2+y} =\int_0^1\left[\log (x^2+y) \right|_{y=0}^1\, dx$$
$$=\int_0^1\left[ \log (1+x^2) -2\log(x)\right] dx $$
$$=\left[ x\log (1+x^2) -2x\log(x)\right|_{x=0}^{1}- 2\int_0^1\left[ \tfrac{x^2}{1+x^2} -1\right] dx$$
$$ =\left[ x\log (1+x^2) -2x\log(x)\right|_{x=0}^{1}- 2\int_0^1\tfrac{1}{1+x^2} dx$$
$$ =\left[ x\log (1+x^2) -2x\log(x)-2\tan ^{-1}(x)\right|_{x=0}^{1} = \tfrac{\pi}{2}+\log 2$$

 

[member 587159]

4) b) $$\int_A \frac{1}{x^2+y}\, d\lambda(x,y) =\int_0^1\int_0^1 \frac{dydx}{x^2+y} =\int_0^1\left[\log (x^2+y) \right|_{y=0}^1\, dx$$
$$=\int_0^1\left[ \log (1+x^2) -2\log(x)\right] dx $$
$$=\left[ x\log (1+x^2) -2x\log(x)\right|_{x=0}^{1}- 2\int_0^1\left[ \tfrac{x^2}{1+x^2} -1\right] dx$$
$$ =\left[ x\log (1+x^2) -2x\log(x)\right|_{x=0}^{1}- 2\int_0^1\tfrac{1}{1+x^2} dx$$
$$ =\left[ x\log (1+x^2) -2x\log(x)-2\tan ^{-1}(x)\right|_{x=0}^{1} = \tfrac{\pi}{2}+\log 2$$

Hi. I think you should justify the first equality. You are using a theorem for this. Which theorem and why can you apply it?

 

[benorin]

If the Improper Riemann and Lebesgue integrals both exist, they are equal?

 

[fresh_42]

If the Improper Riemann and Lebesgue integrals both exist, they are equal?

No.

 

[benorin]

I thumbed through baby Rudin, and alas, no joy. Rudin doesn't distinguish betwixt $d\lambda (x,y)$ and $d\lambda (x)\, d\lambda (y)$, everything is just $d\mu$ no matter the dimension.

 

[fresh_42]

It is neither about the Lebesgue measure nor the dimension. It has to do with the order of integration, something you always have to consider, if you integrate over an area or a volume, because you can't do it in a single step.

 

[member 587159]

If the Improper Riemann and Lebesgue integrals both exist, they are equal?

The problem is that in order to calculate the exact value, like you did, you need to somehow show that the integral exists, which is the original question.

So basically, you assumed what the exercise asks you to prove to get your solution, which seems problematic to me.

I thumbed through baby Rudin, and alas, no joy. Rudin doesn't distinguish betwixt $d\lambda (x,y)$ and $d\lambda (x)\, d\lambda (y)$, everything is just $d\mu$ no matter the dimension.

Baby Rudin does not treat product measures, so you won't get the relevant stuff in that book. See Papa Rudin for a treatment of this instead.

 

[benorin]

7. a) Determine $\int_1^\infty \frac{\log(x)}{x^3}\,dx\,.$
7. b) Determine for which $\alpha$ the integral $\int_0^\infty x^2\exp(-\alpha x)\,dx$ converges.
7. c) Find a sequence of functions $f_n\, : \,\mathbb{R}\longrightarrow \mathbb{R}\, , \,n\in \mathbb{N}$ such that $$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx \neq \int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx$$
7. d) Find a family of functions $f_r\, : \,\mathbb{R}^+\longrightarrow \mathbb{R}\, , \,r\in \mathbb{R}$ such that
$$
\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx \neq \int_\mathbb{R} \lim_{r \to 0} f_r(x) \,dx
$$
7. e) Find an example for which
$$
\dfrac{d}{dx}\int_\mathbb{R}f(x,y)\,dy \neq \int_\mathbb{R}\dfrac{\partial}{\partial x}f(x,y)\,dy
$$
(by @fresh_42 )

Note: I'm going to post this incomplete answer so you can get started, then come back and do the rest later...

7. a) Let $I_a:=\int_1^\infty \frac{\log(x)}{x^3}\,dx$. Let $x=e^u\Rightarrow dx=e^u du$ hence

$$I_a=\int_0^\infty ue^{-3u}e^u \, du=\int_0^\infty ue^{-2u}\, du=\int_0^\infty \tfrac{v}{2}e^{-v}\, \tfrac{dv}{2}=\tfrac{1}{4}\Gamma (2)=\boxed{\tfrac{1}{4}}$$

7. b) Determine for which $\alpha$ the integral $\int_0^\infty x^2\exp(-\alpha x)\,dx$ converges.
Let $I_{\alpha}:=\int_0^\infty x^2e^{-\alpha x}\,dx$. For $\alpha$ positive, let $u=\alpha x$, then

$$I_{\alpha}=\int_0^\infty \left( \tfrac{u}{\alpha}\right) ^2e^{-u}\,\tfrac{du}{\alpha}=\tfrac{1}{\alpha ^3}\int_0^\infty u^2e^{-u}\,du=\tfrac{1}{\alpha ^3}\Gamma (3)=\tfrac{2}{\alpha ^3}$$

Hence $I_{\alpha}$ converges for positive $\alpha$. $I_{\alpha}$ obviously diverges for $\alpha$ non-positive. Let $\alpha :=a+b i$ then $\left| I_{a+b i}\right| \leq\tfrac{2}{a^3}$ whenever $a>0$. Hence $I_{\alpha}$ converges for $\Re \left[\alpha \right] >0$.

7. c) Find a sequence of functions $f_n\, : \,\mathbb{R}\longrightarrow \mathbb{R}\, , \,n\in \mathbb{N}$ such that $$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx \neq \int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx$$

Define $f_n (x):=\begin{cases}xne^{-\tfrac{1}{2}nx^2} & \text{if } x \geq 0 \\0 & \text{if } x \leq 0\end{cases}$

Aside: Let $|a|<1$. Consider the sequence $\left\{ na^n\right\}$: this sequence trivially converges to zero if $a=0$. For $0<|a|<1$, we may write, with $\theta >0$

$$|a|=\tfrac{1}{1+\theta}\Rightarrow |a^n|=\tfrac{1}{1+\binom{n}{1}\theta +\cdots +\binom{n}{n}\theta ^n}$$

so we have for $n=1, 2, 3,\ldots$,

$$|a^n|<\tfrac{1}{1+\binom{n}{2}\theta ^2}\Rightarrow |na^n|<\tfrac{1\cdot 2}{(n-1)\theta ^2}$$

Thus we have $|na^n|<\epsilon ,$ as soon as, $\tfrac{1\cdot 2}{(n-1)\theta ^2}< \epsilon$
i.e. for every $$n>1+\tfrac{2}{\epsilon \cdot \theta ^2}$$
Hence $na^n\rightarrow 0$ for $|a|<1$. End Aside.

Back to our sequence of functions $\left\{ f_n\right\}$: Let

$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx=\sum_{n=1}^\infty \int_{0}^\infty xne^{-\tfrac{1}{2}nx^2}\,dx=\sum_{n=1}^\infty \int_{0}^\infty u^{0}e^{-u}\,du=\sum_{n=1}^\infty \Gamma (1) \rightarrow +\infty$

By the aside with $a=e^{-\tfrac{1}{2}x^2}<1\forall x>0$, we have

$\int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx=\int_\mathbb{R} 0 \, dx=0$

hence

$$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx \neq \int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx$$.

7. d) Find a family of functions $f_r\, : \,\mathbb{R}^+\longrightarrow \mathbb{R}\, , \,r\in \mathbb{R}$ such that
$$\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx\neq \int_\mathbb{R} \lim_{r \to 0} f_r(x) \,dx$$

Define $f_r(x):=\begin{cases}\left\|\tfrac{1}{r}\right\| x(1-x^2)^{\left\|\tfrac{1}{r}\right\| }& \text{if } 0\leq x \leq 1 \\0 & \text{ otherwise } \end{cases}$

where $\left\| y\right\|:=\text{nint}(y)$. Let n be the integer $n:=\left\| \tfrac{1}{r}\right\|$,

$$\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx=\lim_{r \to 0}\int_{0}^{1}\left\|\tfrac{1}{r}\right\| x(1-x^2)^{\left\|\tfrac{1}{r}\right\| }\, dx$$
$$ =\lim_{n \to \infty}\tfrac{n}{2}\int_{0}^{1}u^{n}\, du =\lim_{n \to \infty}\tfrac{n}{2(n+1)}=\tfrac{1}{2} $$
but
$$ \int_{\mathbb{R}} \lim_{r \to 0} f_r(x) \, dx = \int_{0}^{1} \lim_{r \to 0} \left\| \tfrac{1}{r} \right\| x(1-x^2)^{ \left\| \tfrac{1}{r} \right\|} \, dx $$
$$ = \int_{0}^{1} \lim_{n \to\infty} n x(1-x^2)^{n}\,dx =0 $$
where the limit was evaluated by the aside to problem 7 c). Hence
$$\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx\neq \int_\mathbb{R} \lim_{r \to 0} f_r(x) \,dx$$

 

[fresh_42]

I don't see why you can swap limit and sum in 7.c), i.e. why $\sum_{\mathbb{N}}f_n(x) = 0$. This wasn't what you have shown.

... you know that you can look up the solutions?

 

[benorin]

I don't see why you can swap limit and sum in 7.c), i.e. why $\sum_{\mathbb{N}}f_n(x) = 0$. This wasn't what you have shown.

... you know that you can look up the solutions?

I just spent some time figuring that I was wrong, the $f_n(x)$ in 7 c) should have been $s_n(x):=\sum_n f_n(x)$. Sorry, I'll fix it...

I realize the solutions are posted on or about the 15th of each month but I do this because I am bored, and doing this sharpens my skills.

 

[fresh_42]

... and doing this sharpens my skills.

Only if you are accurate in your arguments. Fubini and the existence of an integral with a limit as boundary require a note to prove that you know what you are doing.

And a sum over something small doesn't need to be zero. The harmonic series is the best example.

 

[benorin]

7. a) Determine $\int_1^\infty \frac{\log(x)}{x^3}\,dx\,.$
7. b) Determine for which $\alpha$ the integral $\int_0^\infty x^2\exp(-\alpha x)\,dx$ converges.
7. c) Find a sequence of functions $f_n\, : \,\mathbb{R}\longrightarrow \mathbb{R}\, , \,n\in \mathbb{N}$ such that $$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx \neq \int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx$$
7. d) Find a family of functions $f_r\, : \,\mathbb{R}^+\longrightarrow \mathbb{R}\, , \,r\in \mathbb{R}$ such that
$$
\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx \neq \int_\mathbb{R} \lim_{r \to 0} f_r(x) \,dx
$$
7. e) Find an example for which
$$
\dfrac{d}{dx}\int_\mathbb{R}f(x,y)\,dy \neq \int_\mathbb{R}\dfrac{\partial}{\partial x}f(x,y)\,dy
$$
(by @fresh_42 )

7. a) Let $I_a:=\int_1^\infty \frac{\log(x)}{x^3}\,dx$. Let $x=e^u\Rightarrow dx=e^u du$ hence

$$I_a=\int_0^\infty ue^{-3u}e^u \, du=\int_0^\infty ue^{-2u}\, du=\int_0^\infty \tfrac{v}{2}e^{-v}\, \tfrac{dv}{2}=\tfrac{1}{4}\Gamma (2)=\boxed{\tfrac{1}{4}}$$

7. b) Determine for which $\alpha$ the integral $\int_0^\infty x^2\exp(-\alpha x)\,dx$ converges.
Let $I_{\alpha}:=\int_0^\infty x^2e^{-\alpha x}\,dx$. For $\alpha$ positive, let $u=\alpha x$, then

$$I_{\alpha}=\int_0^\infty \left( \tfrac{u}{\alpha}\right) ^2e^{-u}\,\tfrac{du}{\alpha}=\tfrac{1}{\alpha ^3}\int_0^\infty u^2e^{-u}\,du=\tfrac{1}{\alpha ^3}\Gamma (3)=\tfrac{2}{\alpha ^3}$$

Hence $I_{\alpha}$ converges for positive $\alpha$. $I_{\alpha}$ obviously diverges for $\alpha$ non-positive. Let $\alpha :=a+b i$ then $\left| I_{a+b i}\right| \leq\tfrac{2}{a^3}$ whenever $a>0$. Hence $I_{\alpha}$ converges for $\Re \left[\alpha \right] >0$.

7. c) Find a sequence of functions $f_n\, : \,\mathbb{R}\longrightarrow \mathbb{R}\, , \,n\in \mathbb{N}$ such that $$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx \neq \int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx$$f

Define $f_0(x):=\begin{cases}xe^{-\tfrac{1}{2}x^2}& \text{if } x\geq 0 \\0 & \text{ otherwise }\end{cases}$
and
$f_n (x) := \begin{cases} x e^{-\tfrac{1}{2} x^2} + \sum_{k=2}^{n} \left[ x k e^{-\tfrac{1}{2} k x^2}-x (k-1) e^{-\tfrac{1}{2} (k-1) x^2}\right] & \text{ if } x\geq 0 \\0 & \text{ otherwise} \end{cases}$

Aside: Let $|a|<1$. Consider the sequence $\left\{ na^n\right\}$: this sequence trivially converges to zero if $a=0$. For $0<|a|<1$, we may write, with $\theta >0$

$$|a|=\tfrac{1}{1+\theta}\Rightarrow |a^n|=\tfrac{1}{1+\binom{n}{1}\theta +\cdots +\binom{n}{n}\theta ^n}$$

so we have for $n=1, 2, 3,\ldots$,

$$|a^n|<\tfrac{1}{1+\binom{n}{2}\theta ^2}\Rightarrow |na^n|<\tfrac{1\cdot 2}{(n-1)\theta ^2}$$

Thus we have $|na^n|<\epsilon ,$ as soon as, $\tfrac{1\cdot 2}{(n-1)\theta ^2}< \epsilon$
i.e. for every $$n>1+\tfrac{2}{\epsilon \cdot \theta ^2}$$
Hence $na^n\rightarrow 0$ for $|a|<1$. End Aside.

Back to our sequence of functions $\left\{ f_n\right\}$: Let $s_n(x):=\sum_{k=1}^n f_k(x) = xne^{-\tfrac{1}{2}nx^2}$

$$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx$$
$$=\sum_{n=1}^\infty \int_{0}^\infty\left\{ xe^{-\tfrac{1}{2}x^2} +x\sum_{k=2}^{n} \left[ ke^{-\tfrac{1}{2}kx^2}-(k-1)e^{-\tfrac{1}{2}(k-1)x^2}\right] \right\} \,dx$$
$$=\sum_{n=1}^\infty\left\{ \int_{0}^\infty xe^{-\tfrac{1}{2}x^2}dx +\sum_{k=2}^{n} \left[ \int_{0}^{\infty} xke^{-\tfrac{1}{2}kx^2}dx-\int_{0}^\infty x(k-1)e^{-\tfrac{1}{2}(k-1)x^2}dx\right] \right\}$$
$$=\sum_{n=1}^\infty\left\{ \tfrac{1}{2}\int_{0}^\infty e^{-u_1}du_1 +\sum_{k=2}^{n} \left[ \int_{0}^{\infty} e^{-u_2}du_2-\int_{0}^\infty e^{-u_3}du_3\right] \right\}$$
$$=\sum_{n=1}^\infty\left\{ \tfrac{1}{2}\Gamma (1) +\sum_{k=2}^{n} \left[ \Gamma(1)-\Gamma (1)\right] \right\} =\tfrac{1}{2}\sum_{n=1}^\infty \Gamma (1) \rightarrow +\infty$$

By the aside with $a=e^{-\tfrac{1}{2}x^2}<1\forall x>0$, we have

$\int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx=\int_\mathbb{R}\left(\lim_{n\to\infty}s_n(x) \right) \,dx=\int_\mathbb{R} 0 \, dx=0$

hence

$$\sum_\mathbb{N}\int_\mathbb{R}f_n(x)\,dx \neq \int_\mathbb{R}\left(\sum_\mathbb{N}f_n(x) \right) \,dx$$

7. d) Find a family of functions $f_r\, : \,\mathbb{R}^+\longrightarrow \mathbb{R}\, , \,r\in \mathbb{R}$ such that
$$\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx\neq \int_\mathbb{R} \lim_{r \to 0} f_r(x) \,dx$$

Define $f_r(x):=\begin{cases}\left\|\tfrac{1}{r}\right\| x(1-x^2)^{\left\|\tfrac{1}{r}\right\| }& \text{if } 0\leq x \leq 1 \\0 & \text{ otherwise } \end{cases}$

where $\left\| y\right\|:=\text{nint}(y)$. Let n be the integer $n:=\left\| \tfrac{1}{r}\right\|$,

$$\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx=\lim_{r \to 0}\int_{0}^{1}\left\|\tfrac{1}{r}\right\| x(1-x^2)^{\left\|\tfrac{1}{r}\right\| }\, dx$$
$$ =\lim_{n \to \infty}\tfrac{n}{2}\int_{0}^{1}u^{n}\, du =\lim_{n \to \infty}\tfrac{n}{2(n+1)}=\tfrac{1}{2} $$
but
$$ \int_{\mathbb{R}} \lim_{r \to 0} f_r(x) \, dx = \int_{0}^{1} \lim_{r \to 0} \left\| \tfrac{1}{r} \right\| x(1-x^2)^{ \left\| \tfrac{1}{r} \right\|} \, dx $$
$$ = \int_{0}^{1} \lim_{n \to\infty} n x(1-x^2)^{n}\,dx =0 $$
where the limit was evaluated by the aside to problem 7 c). Hence
$$\lim_{r \to 0}\int_\mathbb{R}f_r(x)\,dx\neq \int_\mathbb{R} \lim_{r \to 0} f_r(x) \,dx$$

7. e) Find an example for which
$$\dfrac{d}{dx}\int_\mathbb{R}f(x,y)\,dy \neq \int_\mathbb{R}\dfrac{\partial}{\partial x}f(x,y)\,dy$$

This is baby Rudin pg. 242 ch 9 exercise #28:

Let $f(x,y)=\begin{cases} y-2\sqrt{-x} & \text{if } \sqrt{|x|}\leq y \leq 2\sqrt{|x|} \wedge x<0 \\ -y & \text{if } 0 \leq y \leq \sqrt{|x|} \wedge x< 0 \\ y & \text{if } 0 \leq y \leq \sqrt{x} \wedge x\geq 0 \\ -y+2\sqrt{x} & \text{if } \sqrt{x}\leq y \leq 2\sqrt{x} \wedge x\geq 0 \\ 0 & \text{ otherwise} \end{cases}$

Let $g(x):=\int_{\mathbb{R}}f(x,y)\, dy=\begin{cases} \int_{\sqrt{-x}}^{2\sqrt{-x}}( y-2\sqrt{-x})\, dy-\int_{0}^{\sqrt{-x}}y\, dy & \text{if } x<0 \\ \int_{0}^{\sqrt{x}}y\, dy+\int_{\sqrt{x}}^{2\sqrt{x}}(-y+2\sqrt{x})\, dy & \text{if } x\geq 0 \end{cases}$

Recall the formula for differentiating under the integral sign:

$$\boxed{\frac{d}{dx}\int_{a(x)}^{b(x)} h(x,y)\, dy =\int_{a(x)}^{b(x)} \tfrac{\partial h}{\partial x}\, dy + h(x,b(x)\tfrac{db}{dx}-h(x,a(x)\tfrac{da}{dx}}$$

Then, differentiating under the integral sign, we have

$$g^{\prime}(x)=
\begin{cases} \int_{\sqrt{-x}}^{2\sqrt{-x}}\left( y+\tfrac{1}{\sqrt{-x}}\right)\, dy+0-(-\sqrt{-x})\left( -\tfrac{1}{\sqrt{-x}}\right) -0-(\sqrt{-x})\left( -\tfrac{1}{2\sqrt{-x}}\right)+0 & \text{if } x<0 \\ 0+(\sqrt{x})\left( \tfrac{1}{2\sqrt{x}}\right) -0+\int_{\sqrt{x}}^{2\sqrt{x}}\left( -y+\tfrac{1}{\sqrt{x}}\right) \, dy +0-(\sqrt{x})\left( \tfrac{1}{2\sqrt{x}}\right) & \text{if } x\geq 0 \end{cases}$$
cc
$$\Rightarrow g^{\prime}(x)=\begin{cases} \left[ \tfrac{1}{2}y^2+\tfrac{y}{\sqrt{-x}}\right|_{y=\sqrt{-x}}^{2\sqrt{-x}} & \text{if } x<0 \\ \left[ -\tfrac{1}{2}y^2+\tfrac{y}{\sqrt{x}}\right|_{y=\sqrt{x}}^{2\sqrt{x}}& \text{if } x\geq 0 \end{cases}=\begin{cases} \tfrac{3}{2}|x|+1 & \text{if } x<0 \\ -\tfrac{3}{2}x+1& \text{if } x\geq 0 \end{cases}=1-\tfrac{3}{2}x$$
$$\Rightarrow \boxed{g^{\prime}(x)=1-\tfrac{3}{2}x, \, \, \forall x\in\mathbb{R}}$$

Now $f_{x}(x,y)=\begin{cases} \tfrac{1}{\sqrt{-x}} & \text{if } \sqrt{|x|}\leq y \leq 2\sqrt{|x|} \wedge x<0 \\ 0 & \text{if } 0 \leq y \leq \sqrt{|x|} \wedge x< 0 \\ 0 & \text{if } 0 \leq y \leq \sqrt{x} \wedge x\geq 0 \\ \tfrac{1}{\sqrt{x}} & \text{if } \sqrt{x}\leq y \leq 2\sqrt{x} \wedge x\geq 0 \\ 0 & \text{ otherwise} \end{cases}$

Hence

$$F(x):=\int_{\mathbb{R}}f_{x}(x,y)\, dy =\begin{cases} \int_{\sqrt{-x}}^{\sqrt{-x}}\tfrac{dy}{\sqrt{-x}}+0 & \text{if } x<0 \\ 0+ \int_{\sqrt{x}}^{\sqrt{x}}\tfrac{dy}{\sqrt{x}} & \text{if } x<0 \\ 0 & \text{if } x=0 \end{cases}$$
$$\boxed{F(x)=\begin{cases} 1 & \text{if } |x|>0 \\ 0 & \text{if } x=0 \end{cases}}$$

thus indeed

$$F(x):=\int_{\mathbb{R}}f_{x}(x,y)\, dy \neq \tfrac{d}{dx}\int_{\mathbb{R}}f(x,y)\, dy =: g^{\prime}(x)$$

 

[benorin]

I fixed 7c) above @fresh_42

 

[fresh_42]

I fixed 7c) above @fresh_42

Again, I don't see that your sum is zero. If it goes to zero at infinity, there is still something to add (integrate) for small numbers. The sum itself isn't zero. At least I don't see it.

 

[benorin]

The definition of $f_n(x)$ is telescopic, giving the nth partial sum $s_n(x):=\sum_{k=1}^n f_k(x) = xne^{-\tfrac{1}{2}nx^2}\rightarrow 0$ as $n\to\infty$, clear?

 

[fresh_42]

No, clear is something else. You should have described this idea and demonstrated that it works. But the idea is correct. However, you don't need such a complicated function. A simple step does the same: "mass vanishes to infinity".

I realize the solutions are posted on or about the 15th of each month ...

No, I meant that I've published my solutions here:
https://www.physicsforums.com/threads/solution-manuals-for-the-math-challenges.977057/

And I have taken many known theorems, inequalities etc. so it is worth trying! Mostly.

 

[benorin]

4) b) $$\int_A \frac{1}{x^2+y}\, d\lambda(x,y) =\int_0^1\int_0^1 \frac{dydx}{x^2+y} =\int_0^1\left[\log (x^2+y) \right|_{y=0}^1\, dx$$
$$\vdots$$
$$ = \tfrac{\pi}{2}+\log 2$$

Let $I_{( a , b )}:=\int_{( a , b ) ^2} \frac{1}{x^2+y^2}\, d\lambda(x,y)$.
Clearly $J:=\int_A \frac{1}{x^2+y}\, d\lambda (x,y) \leq I_{(0,1)}$.

Also, let $B:=\left( -\tfrac{1}{2}, \tfrac{1}{2}\right) ^2$, then

$$J\leq I_{(0,1)}=\int_{B}\frac{d\lambda (x,y)}{\left(x-\tfrac{1}{2}\right) ^2+\left(y-\tfrac{1}{2}\right) ^2}=\int_{B}\frac{d\lambda (x,y)}{\left(x+\tfrac{1}{2}\right) ^2+\left(y+\tfrac{1}{2}\right) ^2}$$
$$\Rightarrow 2I_{(0,1)} =\int_{B}\left[ \frac{1}{\left(x-\tfrac{1}{2}\right) ^2+\left(y-\tfrac{1}{2}\right) ^2}+\frac{1}{\left(x+\tfrac{1}{2}\right) ^2+\left(y+\tfrac{1}{2}\right) ^2}\right] d\lambda (x,y) $$
$$= \int_{B} \frac{4(2x^2+2y^2+1)}{\left(2x^2+2y^2+2x+2y+1\right) \left(2x^2+2y^2-2x-2y+1\right) } d\lambda (x,y)$$
$$\leq \int_{B} \frac{4(2x^2+2y^2+1)}{\left(2x^2+2y^2+1\right) \left(2x^2+2y^2-2x-2y+1\right) } d\lambda (x,y)$$
$$= \int_{B} \frac{4}{\left(2x^2+2y^2+2x+2y+1\right) } d\lambda (x,y)$$
$$\leq \int_{B} \frac{4}{\left(2x^2+2y^2+1\right) } d\lambda (x,y)$$

and since $(x,y)\in B\Rightarrow 1\leq 2x^2+2y^2+1\leq 2$ we may extend the inequality chain further to

$$J\leq \int_{B} \frac{4\left(2x^2+2y^2+1\right) }{\left(2x^2+2y^2+1\right) } d\lambda (x,y) =4\lambda (B) = 4 <+\infty$$

hence $J:=\int_A \frac{1}{x^2+y}\, d\lambda (x,y)$ exists and is finite, and therefore by the theorem of Fubini and the previous post $J=\tfrac{\pi}{2}+\log 2\approx 2.26394\ldots$.

Note: @fresh_42 how did you get off so easy in the solutions? You said "By Fubini... *straight to double integral*" w/o proving it was finite, probably I just did way more than I needed to...

 

[member 587159]

Let $I_{( a , b )}:=\int_{( a , b ) ^2} \frac{1}{x^2+y^2}\, d\lambda(x,y)$.
Clearly $J:=\int_A \frac{1}{x^2+y}\, d\lambda (x,y) \leq I_{(0,1)}$.

Also, let $B:=\left( -\tfrac{1}{2}, \tfrac{1}{2}\right) ^2$, then

$$J\leq I_{(0,1)}=\int_{B}\frac{d\lambda (x,y)}{\left(x-\tfrac{1}{2}\right) ^2+\left(y-\tfrac{1}{2}\right) ^2}=\int_{B}\frac{d\lambda (x,y)}{\left(x+\tfrac{1}{2}\right) ^2+\left(y+\tfrac{1}{2}\right) ^2}$$
$$\Rightarrow 2I_{(0,1)} =\int_{B}\left[ \frac{1}{\left(x-\tfrac{1}{2}\right) ^2+\left(y-\tfrac{1}{2}\right) ^2}+\frac{1}{\left(x+\tfrac{1}{2}\right) ^2+\left(y+\tfrac{1}{2}\right) ^2}\right] d\lambda (x,y) $$
$$= \int_{B} \frac{4(2x^2+2y^2+1)}{\left(2x^2+2y^2+2x+2y+1\right) \left(2x^2+2y^2-2x-2y+1\right) } d\lambda (x,y)$$
$$\leq \int_{B} \frac{4(2x^2+2y^2+1)}{\left(2x^2+2y^2+1\right) \left(2x^2+2y^2-2x-2y+1\right) } d\lambda (x,y)$$
$$= \int_{B} \frac{4}{\left(2x^2+2y^2+2x+2y+1\right) } d\lambda (x,y)$$
$$\leq \int_{B} \frac{4}{\left(2x^2+2y^2+1\right) } d\lambda (x,y)$$

and since $(x,y)\in B\Rightarrow 1\leq 2x^2+2y^2+1\leq 2$ we may extend the inequality chain further to

$$J\leq \int_{B} \frac{4\left(2x^2+2y^2+1\right) }{\left(2x^2+2y^2+1\right) } d\lambda (x,y) =4\lambda (B) = 4 <+\infty$$

hence $J:=\int_A \frac{1}{x^2+y}\, d\lambda (x,y)$ exists and is finite, and therefore by the theorem of Fubini and the previous post $J=\tfrac{\pi}{2}+\log 2\approx 2.26394\ldots$.

Note: @fresh_42 how did you get off so easy in the solutions? You said "By Fubini... *straight to double integral*" w/o proving it was finite, probably I just did way more than I needed to...

I didn't attempt to show it is finite, but here is something you can try: Show that

$$\int_A |f(x,y)| \lambda(dx,dy) < \infty$$

On positive functions, you can apply Fubini without trouble so now you are allowed to use it.

 

[fresh_42]

The entire thing is positive, continuous and bounded, so Fubini applies. Mentioning these facts is sufficient. Otherwise we have to state first which version of Fubini we are talking about, so that we can prove the conditions needed for that version. It's only that one has to check as soon as there is a multiple integral. Especially physicists tend to exchange whatever comes along: limits, summations, integrals, differentiation. Those problems you have solved were meant as a reminder that things can be more complicated in general.

 

[benorin]

The entire thing is positive, continuous and bounded, so Fubini applies.

^This.
That is what I should have answered to the Q: "Justify the initial equality by which theorem?" Except for at the origin, a singlet of measure zero. But that the integrand was unbounded thereat seemed to justify a little caution on my part; that and not knowing which Fubini Theorem variant to use. But then you both are experts, @fresh_42 strikes me as a seasoned Analysis prof, or the equivalent in terms of wit.

 

[benorin]

@fresh_42 added 7. e) to post #63 (done)