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Determine whether the integral converges or diverges $$\int_{1}^{\infty} \frac{\log x +\sin x}{\sqrt{x}} dx$$

  1. Considering $$\int \frac{\log x}{\sqrt{x}}dx$$

Let $a = \log x$, $da = \log(e) x^{-1} dx$ . Let $db = x^{-1/2} dx$, $b = 2x^{1/2}$.

$$\int \log x \cdot x^{-1/2} dx= 2x^{1/2} \log(x) - 2 \log(e) \int x^{-1/2} dx = 2 \sqrt{x} \log(x)- 4 \sqrt{x} log(e) $$ So, $$\int_1^{\infty} \frac{\log x}{\sqrt{x}}dx = \lim\limits_{t \rightarrow \infty} \left[2 \sqrt{x} \log(x)- 4 \sqrt{x} log(e) \right]_1^{t} $$

$$\int_1^{\infty} \frac{\log x}{\sqrt{x}}dx = \lim\limits_{t \rightarrow \infty}\left[2 \sqrt{t} \log(t)- 4 \sqrt{t} log(e) +4 \log(e) \right]$$ Here I am unsure how to argument that the limit goes to infinity because of the "competing terms"

  1. Considering $$\int \frac{\sin x}{\sqrt{x}}dx$$ For this one I tried the integration by part with no success.

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Different questions:

a. How can i argument that the first integral approaches infinity therefore non convergent?

b. How should i integrate the 2nd part?

c. If the first part of the integral diverge, can I conclude that the entire integral diverge?

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    $\begingroup$ For large enough $x$, more simply use: $$\ln(x)+\sin(x)\ge\ln(x)-1>1$$ $\endgroup$ – Simply Beautiful Art Aug 1 '17 at 21:18
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Since $$ \int_{e^2}^{\infty} \frac{\log x +\sin x}{\sqrt{x}} dx \ge \int_{e^2}^{\infty} \frac{\log x -1}{\sqrt{x}} dx \ge \int_{e^2}^{\infty} \frac{1}{\sqrt{x}} dx = \left[ 2\sqrt{x} \right]_{e^2}^\infty $$ the integral diverges.

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    $\begingroup$ Because $\log x -1 > 1$ iff $x> e^2$. $\endgroup$ – Clement C. Aug 1 '17 at 22:11

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