I have to study the following improper integral:
$$\int_{1}^{\infty}\frac{a^x}{\sqrt{x}}dx,\,\, a>0$$
I have thought that $a^x$ converges to 0 when $x\to \infty$ if $|a|<1$ so $\forall \epsilon\,\, \exists \,M>0: \forall x>M: |a^x|<\epsilon$.
So:
$$\frac{a^x}{\sqrt{x}}<\frac{\epsilon}{\sqrt{x}}:=g(x)$$
Now the integral of g(x) diverges but this does not help me to conclude the divergence of the initial integral.
How can I proceed?
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axi
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This answers your question: Convergence of $\int_1^{+\infty} x^a e^{bx}, dx$. (Apply it to $\int_1^{\infty}x^{-1/2}e^{x\ln a}dx$.) – Anne Bauval May 15 '23 at 10:29
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@AnneBauval ok thanks! Only a question...if the integrand function does not converge to 0 then can I conclude something yet? – axi May 15 '23 at 10:41
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If you mean in general, the answer is no. But you can of course conclude if the integrand does have a limit which is $\ne0.$ – Anne Bauval May 15 '23 at 11:25