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Although I have known that $\displaystyle\int_0^\infty {{\sin x} \over x} \, dx = {\pi \over 2}$, I have no idea how to work out $\displaystyle\int_0^{ + \infty } {{\cos x} \over x} \, dx$. How can I?

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  • $\begingroup$ @MichaelHardy Please don't add \displaystyle to titles; see here. The title is perfectly readable without it. $\endgroup$
    – epimorphic
    Oct 17, 2014 at 18:52
  • $\begingroup$ @epimorphic : \displaystyle appeared twice in this title before I edited it. After that it appeared only once. ${}\qquad{}$ $\endgroup$ Oct 17, 2014 at 19:01
  • $\begingroup$ @MichaelHardy Ah, sorry, I only read the beginning of the part being edited. I guess the point still stands that it should be removed altogether. $\endgroup$
    – epimorphic
    Oct 17, 2014 at 19:05

3 Answers 3

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Unfortunately, the integral $$ \int_0^\infty\frac{\cos x}{x}\,dx, $$ diverges.

To see this observe that $$ \int_0^{\pi/4}\frac{\cos x}{x}\,dx\ge \frac{\sqrt{2}}{2}\int_0^{\pi/4}\frac{1}{x}=\infty. $$

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$$\int \frac{\cos x}{x} \, dx = \operatorname{Ci}(x) + C,$$ where $\operatorname{Ci}$ is the Cosine Integral.

Your integral does not converge on the interval $[0,\infty)$, since $\operatorname{Ci}(0)=-\infty$ and $\lim_{a \to \infty}\operatorname{Ci}(a)=0$

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The integral does not converge

simply the result is$$\int_0^{\infty} \frac{\cos x}{x} \, dx = \infty$$

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