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$$\lim_{\lambda \to \infty } \, \int_0^{\frac{\pi }{2}} \left| \frac{\sin (\lambda t)}{t}\right| \, dt$$

I know its $\infty$ but how does this grow as $\lambda \to \infty$

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Logarithm in $\lambda$. Hint: rewrite as $$\int_0^{\frac{\pi}2\lambda}\left|\frac{\sin t}t\right|dt$$ It's evident that $\int_0^{\pi\lambda/2}dt/t\sim \log\lambda$. How to connect this with the above integral?

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    There is an absolute sign and the integral does not converge! – 2017-01-05
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    @RajeshDachiraju What matters if it does not converge? – 2017-01-05
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    Note that $|\sin t|\ge1/\sqrt 2$ for $t\in[k\pi+\pi/4,k\pi+3\pi/4]$ and $|\sin t|\le 1$ for any $t$. Direct computation follows, showing that $c\log\lambda<\text{the integral}$\lambda$ sufficiently large. – 2017-01-05
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    Thanks for the answer. I think I hit the bird with this result! – 2017-01-05
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    Can we say same thing when instead of single $sin$ but a sum of sin functions. I mean $\int_0^{\lambda}\left|\frac{A\sin(at) + B\sin(bt) + C\sin(ct)}t\right|dt \sim K\log(\lambda)$ – 2017-01-31