I need to check conditional/absolute convergence of the integral:

$$f(x) = \int_{1}^{\infty}\cos(x^{2})\,\mathrm dx$$

I tried for a long time and I can't understand what I should do.

I know that $\int_{1}^{\infty}\cos(x^{2})\,\mathrm dx$ converges but I can't find if it's conditional convergence or absolute convergence.

  • $\begingroup$ You may wonder why I edited out part of your question; if so please read this part of the FAQ. $\endgroup$ – Lord_Farin Jul 3 '13 at 10:30

The integral does not converge absolutely. Use the substitution $u = x^2$ to show that it's enough to prove that the following integral diverges to $\infty$: $$ \int_1^\infty \left|\frac{\cos(u)}{2\sqrt{u}}\right| \,du $$

Now follow the same steps as in this answer and the fact that $\displaystyle\sum_{n=1}^\infty \frac{1}{\sqrt{n}}$ diverges.


An idea:

$$t:=x^2\implies dx=\frac{dt}{2\sqrt t}\implies \int\limits_0^\infty\cos x^2dx=\int\limits_0^{\sqrt\frac{\pi}2}\cos x^2dx+\int\limits_{\sqrt\frac{\pi}2}^\infty\cos x^2dx=$$

$$=\int\limits_0^{\sqrt\frac{\pi}2}\cos x^2dx+\frac12\int\limits_{\pi/2}^\infty \frac{\cos t}{\sqrt t}dt$$

the first integral above poses no problem, and for the second we do the following:

$$\int_{\pi/2}^\infty\frac{\cos t}{\sqrt t}du=\sum_{n=1}^\infty\int\limits_{(2n-1)\pi/2}^{\frac{(2n+1)\pi}2}\frac{\cos t}{\sqrt t}dt=:\sum_{n=1}^\infty a_{n}$$

Please do note that we got an alternating series (why?) , and now we check what happens with the monotony of the absolute value of the general term sequence integrating, again by substitution:

$$a_{n+1}=\int\limits_{\frac{(2n+1)\pi}2}^{\frac{(2n+3)\pi}2}\frac{\cos t}{\sqrt t}dt\;\;,\;\;u=t-\pi\;,\;\;du=dt\implies$$

$$a_{n+1}=\int\limits_{\frac{(2n-1)\pi}2}^{\frac{(2n+1)\pi}2}\frac{\cos(u+\pi)}{\sqrt{u+\pi}}du=-\int\limits_{\frac{(2n-1)\pi}2}^{\frac{(2n+1)\pi}2}\frac{\cos u}{\sqrt{u+\pi}}du$$

and thus we get that

$$|a_{n+1}|=\left|\;-\int\limits_{\frac{(2n-1)\pi}2}^{\frac{(2n+1)\pi}2}\frac{\cos u}{\sqrt{u+\pi}}du\;\right|\le\left|\;\int\limits_{\frac{(2n-1)\pi}2}^{\frac{(2n+1)\pi}2}\frac{\cos u}{\sqrt u}du\;\right|=|a_n|\;\;\;\;\;\text{(why?)}$$

This shows the above is a Leibniz series and thus it converges...

  • $\begingroup$ I think your post would look more readable if you didn't use \frac in the limits of your integrals. $\endgroup$ – Lord_Farin Jul 3 '13 at 10:31
  • $\begingroup$ Thanks for the advice, @Lord_Farin, but it's that or write things like $\,(2n-1)\pi/2\,$...I'm not sure which one's worse. $\endgroup$ – DonAntonio Jul 3 '13 at 10:56
  • $\begingroup$ My preference is for $(2n-1)\pi/2$, on the sole merit of keeping the expression readable, as it will display one font size larger than $\frac{(2n-1)\pi}2$ (and I take readability as dominant over almost any other consideration). (Minor point: it's a Leibniz series in your last sentence.) $\endgroup$ – Lord_Farin Jul 3 '13 at 11:09
  • $\begingroup$ Thanks, I think I shall change then the integrals' limits. About Leibni(t)z: it seems to be a wave fashion along the years and I got fixed for what I learned back then in undergraduate school: I had lecturers that wrote the name both ways (one of them of german origin). Behold for example here maths.tcd.ie/pub/HistMath/People/Leibniz/RouseBall/… . But I agree that it seems to be more widespread "Leibniz" so I shall change it once and for all...or until I'll forget. Thanks. $\endgroup$ – DonAntonio Jul 3 '13 at 11:29
  • $\begingroup$ Well @Lord_Farin, I changed just one limit but it really doesn't look good, perhaps because of that summatory sign there. I'll leave it as it is so that other people, if they want, can give their opinions. $\endgroup$ – DonAntonio Jul 3 '13 at 11:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.