Can I use the theorem of differentiation under the integral sign for $ F(x,y) = \int_\sqrt{x}^\sqrt{y} \cos (t^2) \, dt $?

Asked Jan 14 '23 at 11:15

Active Jan 14 '23 at 11:34

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Given

$$ F(x,y) = \int_\sqrt{x}^\sqrt{y} \cos (t^2) \, dt $$

i. Determine the set of definition of $F$

ii. Write, where does exist, the gradient of $F$

i. We can easily see that the function $F(x,y)$ is defined for $x \geq 0$, $ y \geq 0$

ii. For this one I was thinking of using the theorem of differentiation under the integral sign since the conditions for the theorem are verified for $x,y > 0$

Is this reasoning correct?

edited Jan 14 '23 at 11:34

asked Jan 14 '23 at 11:15

Neisy Sofía Vadori

4

Why not just utilize the fundamental theorem of calculus? Because the integrand $\cos(t^2)$ does not involve any of the variables $x$ and $y$. – Sangchul Lee Jan 14 '23 at 11:20
You mean to use the chain rule? @SangchulLee – Neisy Sofía Vadori Jan 14 '23 at 11:24
4

Indeed. You may define $G(t)$ as an antiderivative of $\cos(t^2)$ and write $F(x)=G(\sqrt{y})-G(\sqrt{x})$. This should make the analysis on the differentiability of $F$ much straightforward. – Sangchul Lee Jan 14 '23 at 11:26
The whole thing may also be view as a particular case of Leibniz rule. – Anne Bauval Jan 14 '23 at 11:28
In this particular case I would substitute $u=t^2$ to get $$F(x,y)={1\over 2}\int\limits_x^y{\cos u \over \sqrt{u}},du$$ The integrated function is slightly uglier but the endpoints are nicer. – Ryszard Szwarc Jan 14 '23 at 17:48

Can I use the theorem of differentiation under the integral sign for $ F(x,y) = \int_\sqrt{x}^\sqrt{y} \cos (t^2) \, dt $?

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