Using the fact that $\int_{0}^{\infty}e^{-tx}dx = \frac{1}{t}$, for $t >0$ and from Fubini's theorem conclude that $\lim_{n \rightarrow \infty}\int_{0}^{n}\frac{sinx}{x}dx = \frac{\pi}{2}$.
Any hints would be great, thanks :)
Using Fubini's theorem find the value of sine integral
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measure-theory
1 Answers
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Consider $f(x,t) = e^{-tx} \sin t$ over $(0,\infty) \times (0,n)$. We have no trouble using Fubini here.
To get $\int \left( \int e^{-tx} \sin t \, dt\right) dx$, apply integration by parts twice to get the inner integral (you should be familiar with this method).