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


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 $$\frac{1}{x^2+t^2}$$

  • $\begingroup$ 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

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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.