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let $1<t<2$. I need to evaluate

$$\lim_{\epsilon\rightarrow 0^{+}}\int_{0}^{\infty}\,e^{-\epsilon x}\,\frac{|\sin{x}|^{t}} {x^{t-1}}\,dx$$

If $t>2$ one can easily apply the dominated convergence theorem because $$\int_{0}^{\infty}\,e^{-\epsilon x}\,\frac{|\sin{x}|^{t}} {x^{t-1}}\,dx = \int_{0}^{\infty1}\,e^{-\epsilon x}\,\frac{|\sin{x}|^{t}} {x^{t-1}}\,dx+ \int_{1}^{\infty}\,e^{-\epsilon x}\,\frac{|\sin{x}|^{t}} {x^{t-1}}\,dx\\ \leq \int_{0}^{1}\,\frac{|\sin{x}|^{t}} {x^{t-1}}\,dx+ \int_{1}^{\infty}\,\,\frac{1} {x^{t-1}}\,dx\\ \leq \int_{0}^{1}\,x\,dx+ \int_{1}^{\infty}\,\,\frac{1} {x^{t-1}}\,dx<\infty. $$

Now, if we change variables $\epsilon x\rightarrow x$ we get

$$\int_{0}^{\infty}\,e^{-\epsilon x} \,\frac{|\sin{x}|^{t}} {x^{t-1}}\,dx=\epsilon^{t-2}\int_{0}^{\infty}\,e^{-x}\,\frac{|\sin{\frac{x}{\epsilon}}|^{t}} {x^{t-1}}\,dx $$ and $\lim \epsilon^{t-2}=\infty$ when $t<2$.

Does $\lim \int_{0}^{\infty}\,e^{-x}\,\frac{|\sin{\frac{x}{\epsilon}}|^{t}} {x^{t-1}}\,dx$ exist ?

At least, can we show that $\int_{0}^{\infty}\,e^{-x}\,\frac{|\sin{\frac{x}{\epsilon}}|^{t}} {x^{t-1}}\,dx$ is bounded uniformly in $\epsilon$ when $1<t< 2$?

Thanks

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  • $\begingroup$ Another approach could be using the monotone convergence theorem that implies $\lim_{\epsilon\rightarrow 0^{+}}\int_{0}^{\infty}\,e^{-\epsilon x}\,\frac{|\sin{x}|^{t}} {x^{t-1}}\,dx=\int_{0}^{\infty}\,\frac{|\sin{x}|^{t}} {x^{t-1}}\,dx$. The latter integral diverges for $1\leq t<2$. $\endgroup$ – Medo Jun 21 '18 at 4:13
  • $\begingroup$ So why not do that? $\endgroup$ – zhw. Jun 21 '18 at 16:43
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By the monotone convergence theorem $$\lim_{\epsilon\rightarrow 0^{+}}\int_{0}^{\infty}\,e^{-\epsilon x} \,\frac{|\sin{x}|^{t}} {x^{t-1}}\,dx=\int_{0}^{\infty}\,\frac{|\sin{x}|^{t}} {x^{t-1}}\,dx$$ which is a divergent integral because

$$|\sin{x}|^{t}\geq \sin^{2}{x}$$ for all $\;1\leq t\leq 2\;$ and $$\int_{1}^{\infty}\,\frac{\sin^{2}{x}} {x^{t-1}}\,dx=\frac{1}{2}\int_{1}^{\infty}\,\frac{1-\cos{2x}} {x^{t-1}}\,dx.$$

Finally, the integral $\int_{1}^{\infty}\,\frac{1} {x^{t-1}}\,dx$ diverges while the integral $\int_{1}^{\infty}\,\frac{\cos{2x}} {x^{t-1}}\,dx$ converges by the Dirichlet test.

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