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How would one interpret: $$\frac{\mathrm d}{\mathrm dx}\int_0^x (F(y)-F(x))\,\mathrm dy$$ I don't think I can use the fundamental theorem of calculus here, can I?

J. M. ain't a mathematician
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Angada
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  • For this problem you need that $F$ is a differentiable function. If it is, then indeed the fundamental theorem of calculus comes up here, but it's not immediate. Have you tried anything yet? – Ragib Zaman Oct 08 '11 at 14:40
  • @Ragib: F is differentiable. But then I just get F(x)-F(x) = 0 which doesn't seem to make sense. – Angada Oct 08 '11 at 14:41
  • @Ragib: Ok, Mathematica is giving me -x *F'(x) as the answer. but I can't figure out why exactly! – Angada Oct 08 '11 at 14:47
  • See J.M's answer on the correct way to evaluate it. Your error lies in the fact that your integrand depends on $x$. – Ragib Zaman Oct 08 '11 at 14:48
  • @Ragib: I think i get what it's going -- taking the derivative of the integrand wrt x, then integrating over 0 to x. But where does the fundamental theorem fit in? Even though dy = dx, I can't just swap one for another? – Angada Oct 08 '11 at 14:50
  • Angada, the fundamental theorem applies at the time you evaluate $\frac{\mathrm d}{\mathrm dx} \int_0^x F(y)\mathrm dy$... the $y$ is a dummy; it could've been a $t$ and what I did still applies. – J. M. ain't a mathematician Oct 08 '11 at 14:56
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    That's not exactly what J.M. did. The fundamental theorem of calculus states that if $f$ is differentiable, then $ \frac{d}{dx}\int^x_a f(t) dt = f(x) $. Note that the integrand $f(t)$ does not depend on the upper limit of integration, $x$. So to be able to use FTC, J.M. first had to remove the $F(x)$ term out of the integral. Note also that $F(x)$ is held as a constant whilst integrating with respect to $y$. – Ragib Zaman Oct 08 '11 at 14:56
  • @Ragib -- I actually wrote my last comment before I saw JM's answer. But, I still wouldn't have gotten it, so this is very helpful!! Thanks! – Angada Oct 08 '11 at 15:04
  • Angada: as a tiny piece of advice, next time please don't accept answers until you have truly understood them, and are completely satisfied by them. :) – J. M. ain't a mathematician Oct 08 '11 at 15:22
  • @J.M. -- I thought I understood. I followed what you were doing well enough to answer the question, I just didn't see the link with the FTC. Thank you for your great answer! – Angada Oct 08 '11 at 16:11
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    No problem. I'm just telling you that the check mark is an important piece of feedback, IMHO. If you aren't sure/satisfied with an answer you've gotten, it is perfectly acceptable to hold off accepting an answer... – J. M. ain't a mathematician Oct 08 '11 at 16:14
  • @J.M. Thanks for the feedback. I suppose I am a little quick on accepting in general. – Angada Oct 08 '11 at 16:16

1 Answers1

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Certainly doable:

$$\begin{split}\frac{\mathrm d}{\mathrm dx}\int_0^x (F(y)-F(x))\,\mathrm dy&=\frac{\mathrm d}{\mathrm dx}\left(\int_0^x F(y)\mathrm dy-F(x)\int_0^x \,\mathrm dy\right)\\&=\frac{\mathrm d}{\mathrm dx}\left(\int_0^x F(y)\mathrm dy-x\,F(x)\right)\\&=F(x)-x\,F^\prime(x)-F(x)\\&=-x\,F^\prime (x)\end{split}$$

J. M. ain't a mathematician
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    In the general case, it intuitively ought to hold (under appropriate assumptions) that $$\frac{d}{dx}\int_0^x f(x,y)dy=\int_0^x \frac{\partial f}{\partial x} dy + f(x,x)$$ which agrees with this solution for the case where $f$ can be separated into terms depending only on $x$ and $y$. – hmakholm left over Monica Oct 08 '11 at 15:09
  • @HenningMakholm under what conditions it is going to hold? I am solving a similar problem, but the bounds of integration are $(-\infty, \infty)$ – Fazzolini May 05 '14 at 07:31