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Consider $f(x,y)=y^3e^{-y^2x}$ and define $F(y) =\int_0^{\infty}f(x,y)dx$

We have that $F'(0)\not = \int_0^{\infty} \frac{\partial f}{\partial y}(x,0)dx$

in the spoiler there is how I got this, in case I made a mistake there

We calculate $F'(0)$ essentially using Monotone convergence theorem we can show that, for $y\in \mathbb{R}\setminus\{0\} $, $F(y)=y$ moreover $F(0)=0$ so $F'(0)=1$

Now, I want to understand which hypothesis of Theorem 2 at this page does not hold. Instead of the third hypothesys at the link, though, I use this weacker hypothesis, which is still enough:

"For each $b \in \mathbb{R}$, there exists an open interval $b\in J$ and an integrable function over $(0, \infty)$ , $g(x)$ such that $| \frac{\partial f}{\partial y}(x,y)| \leq g(x)$ for every $y\in J$ and $\forall x$"

Now, the first hypothesis certainly holds as $\forall y, \ x\rightarrow f(x,y)$ is integrable $(0,\infty)$ by comparison with $e^{-kx}$ for appropriate positive value of $k$

Moreover $ \frac{\partial f}{\partial y}(x,y)$ exists everywhere...

So is the last hypothesis to be problematic but I can't see how as I can bound $y$ in $J$ and then just use some linear combination of $e^{-kx}$ and $ xe^{-lx}$ for suitable $k,l$ as they are both integrable over $(0,\infty)$...

Thank you very much!

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1 Answer

The theorem says if you can bound $f_y(x,y)$ with an integrable function $g(x)$, that is $$ |f_y(x,y)|\leq |g(x)|, $$ then you can change the order of differentiation and integration.

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yes but I think you can weacken this condition by bounding $f_y$, for each $b \in \mathbb{R}$ on an open interval containing $b$ (and for all values of x)! –  Moritzplatz Dec 14 '12 at 16:13
    
or do you think I can just say that my $f$ does not respect this condition and hence why the theorem doesn't work? it feels like cheating! –  Moritzplatz Dec 14 '12 at 16:50
    
any idea? the theorem I refer to is not the one you quote but more similar to Theorem 3 in this page planetmath.org/DifferentiationUnderIntegralSign.html –  Moritzplatz Dec 14 '12 at 17:31
    
@Moritz See this answer of Qiaochu for a very general statement. –  Willie Wong Feb 21 '13 at 9:41
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