Differentiation and Integration cannot always be swapped

[Delta2]

Well here is my (I hope successful this time ) attempt at 10. e)

We consider the function $f(x,y)=\begin{cases}\dfrac{e^{-x|y|}}{y}, y\neq 0 \\ 0, y=0\end{cases}$.

For this function for $y\neq 0$ it is for any $x$, $\frac{\partial f}{\partial x}=-\frac{|y|}{y}e^{-x|y|}$.

Also it should be clear that due to the infinite discontinuity at the points $(x,0)$ the integral $\int_\mathbb{R}f(x,y)dy$ does not converge hence the left hand side of the condition of 10. e) is infinite (or undefined).

The right hand side of the condition is $\int_\mathbb{R}-\frac{|y|}{y}e^{-x|y|}dy=\int_{-\infty}^0e^{xy}dy+\int_0^{+\infty} -e^{-xy} dy=0$ ,

So the condition is true for this $f$.

 

[fresh_42]

Well here is my (I hope successful this time ) attempt at 10. e)

We consider the function $f(x,y)=\begin{cases}\dfrac{e^{-x|y|}}{y}, y\neq 0 \\ 0, y=0\end{cases}$. For this function for $y\neq 0$ it is for any $x$, $\frac{\partial f}{\partial x}=\frac{|y|}{y}e^{-x|y|}$.

Also it should be clear that due to the infinite discontinuity at the points $(x,0)$ the integral $\int_\mathbb{R}f(x,y)dy$ does not converge hence the left hand side of the condition of 10. e) is infinite (or undefined).

The right hand side of the condition is $\int_\mathbb{R}\frac{|y|}{y}e^{-x|y|}dy=\int_{-\infty}^0-e^{xy}dy+\int_0^{+\infty} e^{-xy} dy=0$ ,

So the condition is true for this $f$.

I was looking for finite values, i.e. existing integrals.

Edit:

However it suffices that the two sides differ at a point, where the differentials are evaluated. This would make the functions different. Your idea was good. My example also works with $g\, : \,t \longmapsto \exp (-t^2)$ and $g(xy)$. But instead of dividing by $y$, I multiplied by $x$ which avoids singularities.

 

[Delta2]

Ok well, I see so you looking for an example where both sides converge. ok understood.

 

[fresh_42]

Ok well, I see so you looking for an example where both sides converge. ok understood.

See my edit. You have almost the right function.

 

[Delta2]

Interesting, my original line of thinking is that f should have some discontinuities in order for the two sides to be different. So you saying there is some f that is continuous in R yet the two sides are different?

Hmmmm,, trying to follow your hint.. does the function $f(x,y)=xe^{-x|y|}$ does the job?

 

[Delta2]

I don't see what your absolute value of $y$ is good for. I had a slightly different function, such that differentiation will leave me a factor $x$ to make it zero at $x=0$.

Er but the integral on the left hand side will not converge if I don't put |y|...

But let me ask you directly, is your function continuous everywhere in R?

 

[Delta2]

If I understood well, your "main" function is $f(x,y)=xe^{-x^2y^2}$ but is that all, you haven't defined any special discontinuities?

 

[fresh_42]

If I understood well, your "main" function is $f(x,y)=xe^{-x^2y^2}$ but is that all, you haven't defined any special discontinuities?

I first consider $t \longrightarrow \exp(-t^2)$ and calculated the integral. Then I used a continuously differentiable function $f\, : \,(x,y) \longmapsto h(x)g(xy)$ to calculate both sides. If you want to follow this path, find $h(x)$ and do the computations! The latter is part of the question, so wordings like "Also it should be clear that" aren't especially helpful.

 

[Delta2]

ok so for 10. e) I will follow the suggestion of @fresh_42 .

First let's consider the function $g(t)=e^{-t^2}$ and the integral $\int_{\mathbb{R}} g(t)dt=\int_{\mathbb{R}} e^{-t^2} dt=\sqrt \pi$

Next we consider the function $f(x,y)=xg(xy)=xe^{-x^2y^2}$

Then $\int_{\mathbb{R}} f(x,y)dy=\int xg(xy)dy$, integration by substitution $t=xy\Rightarrow dt=xdy$ so we get that (for $x\neq 0$)
$\int_{\mathbb{R}} f(x,y)dy=\int xg(xy)dy=\int g(t)dt=\sqrt\pi$, and the derivative of this with respect to x is zero, so the LHS of the condition is 0. (For $x=0$ this integral is 0).

For the RHS of the condition we first note that $\frac{\partial f}{\partial x}=g(xy)-2x^2y^2g(xy)$, and integrating w.r.t y we get (for $x\neq 0$)
$\int_{\mathbb{R}}\frac{\partial f}{\partial x}dy=\int_{\mathbb{R}} g(xy) dy-2 \int_{\mathbb{R}}x^2y^2g(xy)dy=\frac{\sqrt\pi}{x}-2x^2\frac{\sqrt\pi}{2x^3}$. This is also zero. (For $x=0$ this integral is $\int_{\mathbb{R}}(g(0y)-0)dy=+\infty$).

Well it seems that again I am stumbled upon a non convergence, hmmm, are you sure @fresh_42 that there is function such that both sides converge (for any $x$) yet they differ?

 

[fresh_42]

ok so for 10. e) I will follow the suggestion of @fresh_42 .

That wasn't my suggestion! My suggestion is $f(x,y)=x|x|g(xy)$ with $g(t)=\exp(-t^2)$ which leads to
$$
\sqrt{\pi}=\left.\dfrac{d}{dx}\right|_{x=0} F(x)=\left.\dfrac{d}{dx}\right|_{x=0}\int_\mathbb{R}f(x,y)\,dy \neq \int_\mathbb{R}\left.\dfrac{\partial}{\partial x}\right|_{x=0}f(x,y)\,dy =0
$$

 

[Keith_McClary]

That wasn't my suggestion! My suggestion is $f(x,y)=x|x|g(xy)$ with $g(t)=\exp(-t^2)$ which leads to
$$
\sqrt{\pi}=\left.\dfrac{d}{dx}\right|_{x=0} F(x)=\left.\dfrac{d}{dx}\right|_{x=0}\int_\mathbb{R}f(x,y)\,dy \neq \int_\mathbb{R}\left.\dfrac{\partial}{\partial x}\right|_{x=0}f(x,y)\,dx =0
$$

You mean $dy$ at the end?
I would quibble that $f(0,y)$ is not defined. I thought of making an example using Cantor functions as discussed here, but then the partial derivatives are only defined "almost everywhere", so that is not rigorous either.

 

[fresh_42]

You mean $dy$ at the end?

Thanks, typo corrected. And $f(0,y)=0\,.$

 

[Delta2]

Hmmm @fresh_42 , if we set $F(x)=\int_{\mathbb{R}}f(x,y)dy$ then $F(0)=0$ and

$F'(0)=\lim_{x\rightarrow 0}\frac{F(x)-F(0)}{x-0}=\lim \frac{F(x)}{x}=lim_{x\rightarrow 0} \frac{\sqrt\pi|x|}{x}$ which doesn't exist...

 

[fresh_42]

Hmmm @fresh_42 , if we set $F(x)=\int_{\mathbb{R}}f(x,y)dy$ then $F(0)=0$ and

$F'(0)=\lim_{x\rightarrow 0}\frac{F(x)-F(0)}{x-0}=\lim \frac{F(x)}{x}=lim_{x\rightarrow 0} \frac{\sqrt\pi|x|}{x}$ which doesn't exist...

$F(x) =x\cdot \sqrt{\pi}$ so $F'(x)=\sqrt{\pi}$ which does exist.

 

[Delta2]

$F(x) =x\cdot \sqrt{\pi}$ so $F'(x)=\sqrt{\pi}$ which does exist.

in post #89 I calculated that $\int_{\mathbb{R}}xg(xy)=\sqrt\pi$ where am I wrong there?

Are you saying that $\int_{\mathbb{R}}|x|g(xy)=\sqrt\pi$?

In my opinion it is $F(x)=|x|\sqrt\pi$

 

[fresh_42]

Find an example for which
$$
\dfrac{d}{dx}\int_\mathbb{R}f(x,y)\,dy \neq \int_\mathbb{R}\dfrac{\partial}{\partial x}f(x,y)\,dy
$$
Proof:
We first define $g(t) = \exp(-t^2)$ and show $I:= \int_\mathbb{R} g(x)\,dx = \sqrt{\pi} \,.$
\begin{align*}
I^2&=\left(\int_\mathbb{R}g(x)\,dx \right)\cdot \left(\int_\mathbb{R}g(y)\,dy \right)\\
&=\int_\mathbb{R}\int_\mathbb{R}g(x)g(y)\,dx\,dy\\
&=\int_\mathbb{R}\int_\mathbb{R}\exp(-x^2-y^2)\\
&=\int_0^{\infty}\int_0^{2\pi}r\exp(-r^2)\,d\varphi \,dr\\
&=2\pi \int_0^\infty r\exp(-r^2)dr\\
&=-\pi\left[\exp(-r^2) \right]_0^\infty\\
&=\pi
\end{align*}
The function $h(x,y)=x\cdot g(xy)$ is continuous, however the parameter integral $H(x)=\int_\mathbb{R}h(x,y)\,dy$ is not:
\begin{align*}
H(x)&=\int_\mathbb{R}x\cdot g(xy)\,dy\\
&=\operatorname{sgn}(x) \int_\mathbb{R}|x|\cdot g(xy)\,dy\\
&=\operatorname{sgn}(x) \int_\mathbb{R} g(t)\,dt \\
&=\operatorname{sgn}(x)\cdot \sqrt{\pi}
\end{align*}
Now let $f\, : \,\mathbb{R}\times \mathbb{R} \longrightarrow \mathbb{R}$ the continuously differentiable function $f(x,y)=x|x|g(xy)$ such that we for $F(x)=\int_\mathbb{R}f(x,y)\,dy$
\begin{align*}
F(x)&=\int_\mathbb{R}x|x|g(xy)\,dy\\
&= x \int_\mathbb{R}|x|g(xy)\,dy\\
&=x \int_\mathbb{R}g(t)dt\\
&=x\sqrt{\pi}\quad\quad \text{ and }\\
F\,'(x)&=\sqrt{\pi}
\end{align*}
Differentiation under the integral gives us
$$
\int_\mathbb{R}\dfrac{\partial}{\partial x}f(x,y)\,dy=\int_\mathbb{R}\left(2|x|g(xy)+x|x|\dfrac{d}{dx}g(xy) \right)\,dy
$$
which vanishes at $x=0\,.$ For $x\neq 0$ we get
\begin{align*}
\int_\mathbb{R}\dfrac{\partial}{\partial x}f(x,y)\,dy&=2\int_\mathbb{R}g(t)\,dt - 2\int_\mathbb{R}t^2g(t)\,dt\\
&= 2\sqrt{\pi}-2\left(\left[-\dfrac{t}{2}\exp(-t^2)\right]_{-\infty}^{\infty} + \dfrac{1}{2}\int_\mathbb{R}g(t)\,dt \right)\\
&=\sqrt{\pi}
\end{align*}
In the end we have
$$
\sqrt{\pi}=\left.\dfrac{d}{dx}\right|_{x=0} F(x)=\left.\dfrac{d}{dx}\right|_{x=0}\int_\mathbb{R}f(x,y)\,dy \neq \int_\mathbb{R}\left.\dfrac{\partial}{\partial x}\right|_{x=0}f(x,y)\,dy =0
$$
and although both sides are identical for $x\neq 0$ they differ at $x=0$ and thus cannot be the same.

 

[Delta2]

Well this is interesting, it seems that $\int_{\mathbb{R}}|x|g(xy)=\int_{\mathbb{R}}xg(xy)=\sqrt\pi$ or where am I wrong in the last equality? Where post #9 goes wrong?

quote from post #9

Then $\int_{\mathbb{R}} f(x,y)dy=\int xg(xy)dy$, integration by substitution $t=xy\Rightarrow dt=xdy$ so we get that (for $x\neq 0$)
$\int_{\mathbb{R}} f(x,y)dy=\int xg(xy)dy=\int g(t)dt=\sqrt\pi$

 

[fresh_42]

The point is that we only have $y$ in $g(xy)$ as variable. This alone determines the sign of $xy$. Therefore, $x$ is treated as a positive constant here, because in $t \longmapsto g(t) = \exp(-t^2)$ we have $g(c\cdot t)= \exp(-c^2t^2)$ and the constant factor at our variable has to be positive in $\dfrac{d}{dy}(xy)=|x|\cdot 1\,.$

 

[Delta2]

Well I admit, wolfram seems to agree with you, it says https://www.wolframalpha.com/input/?i=integral+e^(-x^2y^2)dy+from+-inf+to++inf

$\sqrt {x^2}=|x|$...

 

[fresh_42]

Well this is interesting, it seems that $\int_{\mathbb{R}}|x|g(xy)=\int_{\mathbb{R}}xg(xy)=\sqrt\pi$ or where am I wrong in the last equality? Where post #9 goes wrong?

quote from post #9

​

This is just another function than mine was. And you get a quotient at the end: $\dfrac{1}{x}\text{ or }\dfrac{|x|}{x}$ which my solution avoids.

 

[Delta2]

I am not sure...

But can you tell me where the integration by substitution goes wrong at post #9. I used $t=xy$ and ended up with $\int_R xg(xy)dy=\int_R g(t)dt=\sqrt\pi$#...

Apparently there is a subtle mistake here, I ll try to understand it, thanks

OK I found where the mistake is. If $x<0$ and $t=xy$ then for $y=-\infty$ it is $t=+\infty$ and vice versa, so for $x<0$ it is $\int_{-\infty}^{+\infty}xg(xy)dy=\int_{+\infty}^{-\infty}g(t) dt=-\sqrt\pi$

 

[Delta2]

Thanks all and especially @fresh_42 , the search for this counterexample was an interesting learning experience for me!