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I want to find out if the following integrals converge and if possible, find their values.

$$(a) \int_{1}^{2} \frac{\mathrm dx}{\log x}$$

$$(b) \int_{0}^{+\infty} \left| \frac{\sin x}{x} \right| \mathrm dx$$

For $(a)$ I have $$\int_{1}^{2} \frac{1}{\log x}\mathrm dx = \operatorname{li}\left(x\right) + C$$ which is the logarithmic integral function.

An online function calculator gave me the antiderivative $-\operatorname{\Gamma}\left(0,-\ln\left(x\right)\right)$ and apparently it diverges. I don't understand what is meant with this antiderivative here: How does one get that, and why can the value of this integral not be calculated?

Regarding $(b)$ I used an online calculator as well, which gave me

$$\int_{0}^{+\infty} \left| \frac{\sin x}{x} \right| \mathrm dx= -\dfrac{\mathrm{i}\operatorname{\Gamma}\left(0,\mathrm{i}x\right)-\mathrm{i}\operatorname{\Gamma}\left(0,-\mathrm{i}x\right)}{2}$$ and the value ${\pi}/{2}$, but again I don't know if that is correct and how to get that antiderivative.

Can somebody maybe explain that?

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For the first integral, make the substitution $x\mapsto e^x$ and observe that it becomes $$\int_0^{\ln(2)} \frac{e^x}{x}dx$$ You may use the fact that $\int_0^1 dx/x$ diverges to prove the divergence of this integral.

As for the second integral: express it as a sum $$\int_0^\infty \bigg|\frac{\sin(x)}{x}\bigg|dx=\sum_{n=0}^\infty \int_{\pi n}^{\pi (n+1)}\bigg|\frac{\sin(x)}{x}\bigg|dx$$ Then notice that $$\int_{\pi n}^{\pi (n+1)}\bigg|\frac{\sin(x)}{x}\bigg|dx\gt \int_{\pi n}^{\pi (n+1)}\bigg|\frac{\sin(x)}{\pi(n+1)}\bigg|dx=\frac{2}{\pi}\frac{1}{n+1}$$ and use the divergence of the harmonic series and the comparison test to finish.

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  • $\begingroup$ Hi, thanks for your answer. Can you evaluate how its done with the divergence of the harmonic series and the comparison test? Thanks :) $\endgroup$ – Math Dummy Dec 4 '18 at 6:57
  • $\begingroup$ Well, since each integral is greater than a constant times $\frac{1}{n+1}$, and $\sum \frac{1}{n+1}$ diverges, by the comparison test, the sum of integrals diverges. $\endgroup$ – Franklin Pezzuti Dyer Dec 4 '18 at 14:27

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