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Why is it wrong? $$ \frac{d^2}{dx^2}\int_{-1}^1\log|x-t|dt=\int_{-1}^1\frac{\partial^2}{\partial x^2}\log|x-t|dt=\int_{-1}^1\frac{-1}{(x-t)^2}dt. $$

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What is the domain of $x$? –  Lukas Geyer Oct 27 '12 at 21:00
    
No information about the domain of $x$. –  Sam Oct 27 '12 at 21:04
    
what should the answer be? –  john mangual Oct 27 '12 at 21:06
    
Want to show why I cannot exchange the order of differentiation and integral. –  Sam Oct 27 '12 at 21:12
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If $x\in [-1,1]$, the integral diverges already for the first derivative. If $x$ is outside this interval, I don't see anything wrong with this. –  Lukas Geyer Oct 27 '12 at 21:19

2 Answers 2

up vote 1 down vote accepted

Oh boy... let's try defining the second derivative as a difference quotient: $$ f''(x) = \lim_{\epsilon \to 0} \frac{f(x+\epsilon) - 2f(x) + f(x-\epsilon) }{\epsilon^2} \text{ where } f(x) = \int_{-1}^1 \log |x-t| dt $$ Now let's try it with your integral: \begin{eqnarray}\frac{d^2}{dx^2} \int_{-1}^1 \log |x-t| dt &=& \lim_{\epsilon \to 0} \frac{1}{\epsilon^2} \int_{-1}^1 \log \left| \frac{(x-t)^2-\epsilon^2 }{(x-t)(x-t)}\right| dt \\ &=& \lim_{\epsilon \to 0} \frac{1}{\epsilon^2} \int_{-1}^1 \log \left|1 - \frac{ \epsilon^2}{(x-t)^2}\right| dt\end{eqnarray}

As long as $|x - t| > \epsilon$ we can use Taylor approximation $\log (1 - x) = - x + O(x^2)$:

$$ \frac{1}{\epsilon^2} \int_{-1}^1 \frac{ -\epsilon^2}{(x-t)^2} dt = \int_{-1}^1 \frac{ -1}{(x-t)^2} dt $$ So $f(x)$ is divergent whenever $x \in [-1,1]$.


In that case, you may wish to consider $f(x \pm i\delta)$

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I think john's proof is sufficient to show that I cannot exchange the order of differentiation and integral. Thank you very much. –  Sam Oct 28 '12 at 3:21

Setting $$ I(x):=\int_{-1}^1\log|x-t|dt, $$ we have $$ I(\pm1)=2\log2-2, $$ and for every $x \in \mathbb{R}\setminus\{-1,1\}$: $$ I(x)=\left[(t-x)\log|t-x|-t\right]_{-1}^1=(1-x)\log|1-x|+(1+x)\log|1+x|-2. $$ Clearly $$ I''(x)=\int_{-1}^1\frac{\partial^2}{\partial x^2}\log|x-t|dt \quad \forall x \in \mathbb{R}\setminus\{-1,1\}. $$ Furthermore $I$ is not differentiable at $x=\pm1$, and therefore $I''(\pm1)$ does not exist.


Added For every $x \in \mathbb{R}\setminus\{-1,1\}$ we have \begin{eqnarray} I'(x)&=&\log|1+x|-\log|1-x|=\int_{-1}^1\frac{dt}{x-t}dt=\int_{-1}^1\frac{\partial}{\partial x}\log|x-t|dt\cr I''(x)&=&\frac{1}{1+x}+\frac{1}{1-x}=\int_{-1}^1\frac{-1}{(x-t)^2}dt=\int_{-1}^1\left(\frac{\partial^2}{\partial x^2}\log|x-t|\right)dt. \end{eqnarray} Now the cases $x=1$ and $x=-1$. We just treat the case $x=1$, the other one (i.e. $x=-1$) can be treated in the same manner.

For every $h \ne 0$ we have \begin{eqnarray} \frac{I(1+h)-I(1)}{h} &=&\int_{-1}^1\frac{\log|1-t+h|-\log|1-t|}{h}dt =\int_0^2\frac{\log|s+h|-\log|s|}{h}ds\cr &=&\frac{1}{h}\left[(s+h)\log|s+h|-s\log|s|\right]_0^2\cr &=&\frac{(2+h)\log|2+h|-2\log2}{h}-\log|h|. \end{eqnarray} It follows that $$ \lim_{h \to 0}\frac{I(1+h)-I(1)}{h}=\infty $$

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Thank you very much, Mercy and john, for your prompt responses. I just need time to undertstand your answers. I notice there is a difference between your conclusions. Mercy said the equalities in the problem are fine except at -1 and 1; john said it is fine except [-1,1]. I tested them by x=0, and found it is not divergent and both get the reult 2. Thus, I think probably Mercy is correct. –  Sam Oct 28 '12 at 1:08

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