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We know for uniformly continuous functions in an interval [a.b] Leibniz rule applies $$\begin{aligned} \frac{d}{d\alpha}\int_{a(\alpha)}^{b(\alpha)} f(x,\alpha)\,dx &= \frac{d b(\alpha)}{d \alpha}\,f(b(\alpha),\alpha)-\frac{d a(\alpha)}{d \alpha}\,f(a(\alpha),\alpha)\\ +& \int_{a(\alpha)}^{b(\alpha)}\frac{\partial}{\partial \alpha}\,f(x,\alpha)\,dx\end{aligned} $$

For this integral $$G(x)=\int_0^x\frac{1}{x^2+t^2}dt$$

Differentiate wrt x using Leibniz rule and chain rule gives (labeling the 2nd variable x as x' to prevent confusion)

$$\frac{dG}{dx}=\frac{\partial}{\partial x}\int_0^x\frac{1}{x'^2+t^2}dt\frac{dx}{dx}+\frac{dG}{dx}=\frac{\partial}{\partial x'}\int_0^x\frac{1}{x'^2+t^2}dt$$ $$=\frac{\partial}{\partial x}\int_0^x\frac{1}{x'^2+t^2}dt\frac{dx}{dx}+\int_0^x\frac{\partial}{\partial x'}\left(\frac{1}{x'^2+t^2}\right)dt$$ $$=\frac{1}{x'^2+x^2}+\int_0^x\frac{-2x'}{(x'^2+t^2)^2}dt$$

Identifying x and x' $$=\frac{1}{x^4}+\int_0^x\frac{-2x}{(x^2+t^2)^2}dt$$

which is the same answer as the worked solution of the tutorial exercise where I got this question from

However there was a curious question that arises: Since all the terms in the answers are basically functions of x. I then wonder whether the question itself can be done as a one variable function and solved using chain rule (since Leibniz rule can be derived form chain rule (with the $\frac{\partial}{\partial x}\int$interchange justified by uniform continuity)

So we already have $G(x)$

Now let

$$\Lambda(x)=f(x,x)=\frac{1}{x^4}$$

However I don't know how to deal with these two terms/functions, can they be expressed just as a function of x (since t is just a dummy variable thus play no role)?

$$\int_0^x ()dt$$ and $$\frac{1}{x^2+t^2}$$

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  • $\begingroup$ There seems to be a possible misinterpretation of Leibnitz's rule. Under the integral, $t$ is just a dummy variable while $x$ appears both as a parameter and in the limits of integration. The function $G(x)$ depends on this parameter and this limit of integration. So, upon forming a derivative, one must account for both of these dependencies. Please let me know how the answer I posted can be improved. I really want to help and give you the best answer I can. $\endgroup$ – Mark Viola Apr 17 '15 at 15:35
  • $\begingroup$ Is it because $G(x)$ depends on both the parameter x and integral limit x, so when using the chain rule, there will be two $\frac{\partial}{\partial x}$ terms but the two partial derivatives are technically not the same because of how one x is a parameter while another is part of the integral limit x, so the integrand need to be interpreted as a two variable function and the integral a function of x? $\endgroup$ – Secret Apr 17 '15 at 17:02
  • $\begingroup$ Well, it isn't the chain rule, it is Leibnitz's Rule ... they are not the same. So, think of forming a difference quotient on $G(x)$, with $(G(x+h)-G(x))/h=\frac1h (\int_0^{x+h} f(x+h,t) dt - \int_0^{x} f(x,t) dt) =\int_0^{x+h}\frac1h ( f(x+h,t)-f(x,t)) dt + \frac1h \int_x^{x+h} f(x,t) dt$. Observe that the first term approaches the integral of the $\frac{\partial f}{\partial x}$, while the second term approaches the integrand evaluated at $t=x$. $\endgroup$ – Mark Viola Apr 17 '15 at 17:55
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Here is the correct application of Leibnitz's Rule>

Suppose $G(x)= \int_0^x \frac{dt}{t^2+x^2}$. Then, $G'(x)$ is given by

$$\begin{align} G'(x)&=\left(\frac{1}{t^2+x^2}\right)|_{t=x} +\int_0^x \frac{\partial}{\partial x}\left(\frac{1}{t^2+x^2}\right) dt\\\\ &=\left(\frac{1}{2x^2}\right) +\int_0^x \frac{-2x}{(t^2+x^2)^2}dt\\\\ &\left(\frac{1}{2x^2}\right)-2x\left(\frac{\arctan(t/x)+\frac{xt}{t^2+x^2}}{2x^3}\right)|_0^x\\\\ &=-\frac{\pi/4}{x^2} \end{align}$$

Note that this same result can be obtained by first computing $G(x)$ as

$$G(x)=\int_0^x \frac{dt}{t^2+x^2}=\left(\frac{\arctan(t/x)}{x}\right)|_0^x=\frac{\pi/4}{x}$$

whereupon taking the derivative with respect to $x$ reveals that $G'(x)=-\frac{\pi/4}{x^2}$ as expected!

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