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I am trying to understand what is wrong with the proof posted here that $0=1$ (source):

Given any $x$, we have (by using the substitution $u=x^2/y$) $$\large\int_0^1\frac{x^3}{y^2}e^{-x^2/y}\,dy=\Biggl[xe^{-x^2/y}\Biggr]_0^1=xe^{-x^2}.$$ Therefore, for all $x$, \begin{align} e^{-x^2}(1-2x^2)&=\frac{d}{dx}(xe^{-x^2})\\[0.5em] &= \frac{d}{dx}\int_0^1\frac{x^3}{y^2}e^{-x^2/y}\,dy\\[0.5em] &= \int_0^1\frac{\partial}{\partial x}\Biggl(\frac{x^3}{y^2}e^{-x^2/y}\Biggr)\,dy\\[0.5em] &= \int_0^1 e^{-x^2/y}\Biggl(\frac{3x^2}{y^2}-\frac{2x^4}{y^3}\Biggr)\,dy. \end{align} Now set $x=0$; the left-hand side is $e^0(1-0)=1$, but the right-hand side is $\int_0^1 0\,dy=0$.

Surely it must be incorrect somewhere however I am not sure at which step it fails.

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    $\begingroup$ It's intuitively clear that the bad bit happens when you bring the derivative under the integral sign. Consider the conditions on the proof of that theorem. $\endgroup$ – preferred_anon Jul 26 '15 at 22:00
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    $\begingroup$ Be careful, the integral is not a "proper integral." So there is also a limit thing happening. That is to say, you are bringing the $\frac{d}{dx}$ inside a limit and then inside an integral. Both operations are risky and cannot always be done. $\endgroup$ – Thomas Andrews Jul 26 '15 at 22:03
  • $\begingroup$ See here: math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf pg 13. $\endgroup$ – Moya Jul 26 '15 at 22:50
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Too long for a comment, but not an answer.

The fractions $1/y$ are a red herring, at least.

Let $u=\frac{1}{y}$ then you have:

$$f(x)=\int_{0}^1\frac{x^3}{y^2}e^{-x^2/y}\,dy = \int_{1}^{\infty} x^3e^{-x^2u}\,du=xe^{-x^2}$$

Then the argument would have:

$$\begin{align}e^{-x^2}(1-2x^2)&=f'(x)\\ &=\frac{\partial}{\partial x} \int_{1}^{\infty} x^3e^{-x^2u}\,du \\&=\int_{1}^{\infty}\frac{\partial}{\partial x}(x^3e^{-x^2u})\,du \\&=\int_{1}^{\infty}(3x^2-2x^4u)e^{-x^2u}\,du \end{align}$$

So the same argument works without the fractions.

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  • $\begingroup$ Thank you for this. The fractions were rather jarring. $\endgroup$ – Cameron Williams Jul 26 '15 at 22:29
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    $\begingroup$ The integral $\int_1^\infty x^2 e^{-x^2u}\,du$ does not go to zero when $x\to 0$. $\endgroup$ – A.Γ. Jul 26 '15 at 22:41
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    $\begingroup$ Ah, yes, that's true. $$\int_1^{\infty} e^{-x^2u}\,du =\frac{1}{x^2},$$ so $x^2\int_1^{\infty} e^{-x^2u}\,du$ has value $1$ for all $x>0$. $\endgroup$ – Thomas Andrews Jul 26 '15 at 22:50
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    $\begingroup$ @ThomasAndrews It is $u$ missing in the last $2x^4$, should be $2x^4u$. So the derivation under the integral works ok here, it is the limit $x\to 0$ that fails. $\endgroup$ – A.Γ. Jul 26 '15 at 23:12
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    $\begingroup$ Whoops, yes, corrected. @A.G. $\endgroup$ – Thomas Andrews Jul 26 '15 at 23:17
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Continuing @Thomas Andrews nice calculations that simplify understanding, one can notice that there is no problem here with derivation under the integral sign. It is actually true that $$ e^{-x^2}(1-2x^2)=\int_1^\infty (3x^2-2x^4u)e^{-x^2u}\,du. $$ The problem is that it is not possible to set $x=0$. Looking, for example, at the first part of the last integral $$ \int_1^\infty x^2e^{-x^2u}\,du $$ one could believe that setting $x=0$ would give the zero integrand and, hence, the zero integral. Let us see what happens when $x\to 0$. The integrand, indeed, goes to zero uniformly (that would be enough for a finite measure set), however, for small $x$ the tails of the integrand (for large $u$) become heavier and heavier. If we maximize the integrand for a given $u$ we get $e^{-1}/u$ at $x^2=1/u$. It is not an integrable function, so Lebesgue Dominated Convergence theorem cannot be applied. Neither monotone convergence is true here. In fact, the integral goes to one $$ \int_1^\infty x^2e^{-x^2u}\,du=\int_1^\infty e^{-x^2u}\,d(x^2u)= -e^{-x^2u}\bigg|_1^\infty=e^{-x^2}\to 1, \quad x\to 0. $$

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Let $u=x^2/y$, then $y=x^2/u$: $$ \int_0^1 e^{-x^2/y}\left(\frac{3x^2}{y^2}-\frac{2x^4}{y^3}\right)\,\mathrm{d}y =\int_{x^2}^\infty e^{-u}(3-2u)\,\mathrm{d}u $$ Therefore, $$ \begin{align} \lim_{x\to0}\int_0^1 e^{-x^2/y}\left(\frac{3x^2}{y^2}-\frac{2x^4}{y^3}\right)\,\mathrm{d}y &=\int_0^\infty e^{-u}(3-2u)\,\mathrm{d}u\\[6pt] &=1 \end{align} $$ which is not $0$ as we would seem to get by plugging $x=0$ into the integral on the left. That is, the integral on the left is not continuous at $x=0$.

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