limits and integrals

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Show that $$ \lim_{n \to \infty} \int\limits_{0}^{h} \frac{\sin (n\varepsilon)}{\varepsilon} \;\mathrm{d}\varepsilon = \int\limits_{0}^{\infty} \frac{\sin (t)}{t} \;\mathrm{d}t, \;\;h>0 $$

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Hint: Let $t = n\epsilon $. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$