Show that ∫10 (sin(1/x))/xn 𝑑𝑥 ; 𝑥 > 0 convergence absolutely, if 𝑛 < 1.
"u=1\/x, du=-dx\/x^2"
"\u222b^1_0 (sin(1\/x))\/x^n \ud835\udc51\ud835\udc65=\u222b^{\\infin}_0 (sin(u))u^{n-2} \ud835\udc51u"
since n < 1:
"|(sin(u))u^{n-2}|\\le 1\/u^2"
since series "\\sum 1\/u^2" converge, then "\\sum (sin(u))u^{n-2}" converge absolutely.
So,
"\u222b^1_0 (sin(1\/x))\/x^n \ud835\udc51\ud835\udc65=\u222b^{\\infin}_0 (sin(u))u^{n-2} \ud835\udc51u" convergences absolutely, if 𝑛 < 1.
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