I am trying to find values of $p \in \mathbb{R}$ such that $\displaystyle\int_0^{+\infty} x^p\sin(e^x)$ converges. All I have managed doing is using reduction formulas but I couldn't reach a result. Any ideas?
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Near $+\infty$ one has by partial integration \begin{equation} \int x^p \sin(e^x) d x = -x^p e^{-x} \cos(e^x) + \int(p x^{p-1}-x^p)e^{-x}\cos(e^x) d x \end{equation} which clearly converges for all $p\in {\mathbb R}$
Near $0$, we have $x^p \sin(e^x) \sim x^p \sin(1)$ which converges iff $p> -1$
Gribouillis
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Where does the $e^{-x}$ comes from ? – EDX Aug 03 '20 at 10:21
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@EDX I write $x^p \sin(e^x) = x^p e^{-x} e^x \sin(e^x)$ because I want to use that $e^x \sin(e^x) = (-\cos(e^x))'$ – Gribouillis Aug 03 '20 at 14:25
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Ok nice thanks :) – EDX Aug 03 '20 at 16:13