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- Homework Statement
- Determine the range of ##f(x)## for ##x>0## using differentiation under the integral sign.

- Relevant Equations
- ##f(x)=\int_x^{2x} (\frac{e^{-t^2x}}{t}) \ dt ## for ##x>0##.

The strategy here would probably be to find a differential equation that ##f## satisfies, but differentiating with respect to ##x## using Leibniz rule yields

Continuing to differentiate will yield the integral term again (with ##t^3## instead of ##-t##) together with the derivative of the second term above. Is this the right strategy?

As a bonus question. Consider the function ##4t^2e^{-t^2}\cos{(2xt)}##. This is the second derivative w.r.t. ##x## of the function ##e^{-t^2}\cos{(2xt)}##. Now see the attached image. How does one find a function ##g(t)## that satisfies the inequality in the theorem stated in order to justify the differentiation under the integral sign? (##t## ranges from ##0## to ##\infty## and ##x\in\textbf{R}##)

##f'=\int_x^{2x} (-te^{-t^2x}) \ dt + \frac{2e^{-4x^3}-e^{-x^3}}{x}##

Continuing to differentiate will yield the integral term again (with ##t^3## instead of ##-t##) together with the derivative of the second term above. Is this the right strategy?

As a bonus question. Consider the function ##4t^2e^{-t^2}\cos{(2xt)}##. This is the second derivative w.r.t. ##x## of the function ##e^{-t^2}\cos{(2xt)}##. Now see the attached image. How does one find a function ##g(t)## that satisfies the inequality in the theorem stated in order to justify the differentiation under the integral sign? (##t## ranges from ##0## to ##\infty## and ##x\in\textbf{R}##)

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