Can you prove the following improper integral is convergent?
$$ \int_0^{1/2}\frac{\sin x}{x}\frac{1}{\log\frac{1}{x}}\bigg(1+\frac{1}{N}\log\frac{1}{x}\bigg)^N\,dx. $$
Can you prove the following improper integral is convergent?
$$ \int_0^{1/2}\frac{\sin x}{x}\frac{1}{\log\frac{1}{x}}\bigg(1+\frac{1}{N}\log\frac{1}{x}\bigg)^N\,dx. $$
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Hint. As $x \to 0^+$, one has $$ \frac{\sin x}{x}\frac{1}{\log\frac{1}{x}}\bigg(1+\frac{1}{N}\log\frac{1}{x}\bigg)^N \sim \frac1{N^N}\cdot(-\log x)^{N-1} $$ and the given integral is convergent. One may recall that, for $0<\varepsilon<1/2$, we have $$ \int_0^{\large\varepsilon} (-\log x)^{N-1}dx=\int_{\large -\log \varepsilon}^\infty t^{N-1}e^{-t}dt<\infty $$ for all values of $N$.