Wednesday, 27 March 2019

calculus - Evaluating limbtoinftyintb0fracsinxx,dx=fracpi2











Using the identity lim can I get a hint to show that \lim_{b\to\infty} \int_0^b \frac{\sin x}{x} \,dx= \frac{\pi}{2}.


Answer



Hint:
\begin{align} \lim_{b\to \infty}\int_{0}^{b}\frac{\sin x}{x}dx &= \lim_{a,b\to \infty}\int_{0}^{b}\int_{0}^{a}e^{-xt}dt\sin x dx\\& = \lim_{a,b\to \infty}\int_{0}^{b}dt\int_{0}^{a}e^{-xt}\frac{e^{ix}-e^{-ix}}{2i} dx \\&=\lim_{a,b\to \infty}\int_{0}^{b}dt\int_{0}^{a}\frac{e^{-(t-i)x}-e^{-(i+t)x}}{2i} dx\end{align}.



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