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Suppose $f(x,y)$ is continuous everywhere except where $y=x$. Does FTC apply in a scenario where $x$ is given as a boundary? If so, how? Here's a general example.

$$\frac{d}{dx}\int_x^bf(x,y) dy$$

There is a related post here that doesn't answer this question. The specific integral I'm looking at is more like

$$\frac{d}{dx}\int_x^\infty \frac{g(y)}{\sqrt{y^2-x^2}} dy$$

where it is known that the original integral (prior to differentiation) converges for all $x\ge 0$. I'm tentatively approaching the derivative as

$$\frac{d}{dx}\int_x^c\frac{g(y)}{\sqrt{y^2-x^2}} dy + \lim_{b\to\infty}\int_c^b\frac{d}{dx}\frac{g(y)}{\sqrt{y^2-x^2}} dy$$

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zahbaz
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1 Answers1

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I ended up solving my specific problem by introducing a substitution to set the limits of integration to something nicer.

\begin{align} \frac{d}{dx}\int_x^\infty \frac{g(y)}{\sqrt{y^2-x^2}} dy & =\frac{d}{dx}\int_0^\infty \frac{g(y)}{\sqrt{u}} \frac{du}{2y} &\text{ using: } u = y^2 - x^2 \end{align}

Since $u$ is a dummy variable, I don't see an issue with interchanging the order of operations. From there, one can apply the chain rule, and then undo the substitution.

\begin{align} \frac{d}{dx}\int_0^\infty \frac{1}{\sqrt{u}} \frac{g(y)}{2y} du &= \int_0^\infty \frac{1}{\sqrt{u}} \frac{d}{dx}\left(\frac{g(y)}{2y}\right)du \\ &= \int_0^\infty \frac{1}{\sqrt{u}} \frac{d}{dy}\left(\frac{g(y)}{2y}\right)\frac{dy}{dx} du \\ &= \int_x^\infty \frac{1}{\sqrt{y^2-x^2}} \frac{d}{dy}\left(\frac{g(y)}{2y}\right)\frac{x}{y} (2ydy) \\ &= x\int_x^\infty \frac{1}{\sqrt{y^2-x^2}} \frac{d}{dy}\left(\frac{g(y)}{y}\right) dy \end{align}

So, in conclusion: \begin{align} \frac{d}{dx}\int_x^\infty \frac{g(y)}{\sqrt{y^2-x^2}} dy &= x\int_x^\infty \frac{1}{\sqrt{y^2-x^2}} \frac{d}{dy}\left(\frac{g(y)}{y}\right) dy \end{align}

zahbaz
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  • There shouldn't be an $y$ after you have substituted with $u$... – user251257 Nov 14 '16 at 23:34
  • @user251257 Hmm... I'm not sure if it changes the result. We could write $y= \sqrt{u+x^2}$, but that would not improve readability. My actual problem began with an extra factor of $y$ in the numerator, so it was a bit more pleasant. That is, I was looking at $g(y) = y h(y)$. – zahbaz Nov 14 '16 at 23:49
  • After substitution there is no $y$ anymore. If you are unsure, you should revisit how substation works. – user251257 Nov 14 '16 at 23:52