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I was trying to provide the same answer my mathematics professor gave me when I asked to the problem raised in this thread. What I was told was that instead of summing what we were really doing is taking an integral, so

$x^2 = \sum_1^xx$ becomes $\int_1^xx\,dx$

Then, by the Leibniz integral rule

$\frac d{dx}\int_1^xx\,dx = \int_1^x(\frac d{dx}x)\,dx + x|_x(\frac d{dx}x) -x|_1(\frac d{dx}1) = 2x$

Which is of course $\frac d{dx}x^2$

But, $\int_1^xx\,dx =\frac12x^2-\frac12 \neq x^2$

You could use $x^2=\int_0^x2x\,dx$ but then

$\frac d{dx}\int_0^x2x\,dx = \int_0^x(\frac d{dx}2x)\,dx + 2x|_x(\frac d{dx}x) -2x|_0(\frac d{dx}0) = 4x$

This is in some ways exactly the problem we had before. Is there something wrong about the way I am applying Leibniz's integral law, or is there something more fundamental going on? Was my professor wrong?

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kleineg
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2 Answers2

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Most falsehoods are half-truths; in this case, you forgot something, namely that the number of terms itself is also variable, depending on x: obviously, this significant detail also has to feature in the derivative's final expression. $$\left(\sum_1^xx\right)'_x=\sum_1^xx'+\sum_1^{x'}x=\sum_1^x1+\sum_1^1x=1\cdot x+x=2x.$$

Lucian
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  • That is equation for equation what I brought to my professor as a possible solution. He said that my using that notation was incorrect and that I should use integrals instead. – kleineg Jan 15 '14 at 15:24
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The expression $$ \int_1^xx\,dx $$ is something you should be very careful around. The variable you're integrating over (in this case $x$) should in most cases not appear outside the integrand. I believe what you meant to write is $$ \frac d{dx}\int_1^xx'\,dx' $$ or something to that effect. The Leibniz integral rule does not apply to this setting. The reason is that you're differentiating with respect to the upper limit of the integral, and this is the domain of the fundamental theorem of calculus.

The Leibniz integral rule is mostly used when dealing with functions of multiple variables, and you integrate over one (with constant integrating bounds) while you differentiate with respect to another.

Arthur
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