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I would like to know an alternative proof for the convergence of the improper integral

$$\int_0^\infty\sqrt t\cos(t^2)\mathrm dt$$

The proof that I did was this: with the change of variable $t^2=u$, and because $t\ge 0$, we have that

$$\int_0^\infty\sqrt t\cos(t^2)\mathrm dt=\frac12\int_0^\infty\frac{\cos u}{\sqrt[4]{u}}\mathrm du=\frac12\int_0^{\pi/2}\frac{\cos u}{\sqrt[4]{u}}\mathrm du+\frac12\sum_{k=0}^\infty\int\limits_{\frac\pi2+k\pi}^{\frac{3\pi}2+k\pi}\frac{\cos u}{\sqrt[4]{u}}\mathrm du$$

If we define $$a_k:=\int\limits_{\frac\pi2+k\pi}^{\frac{3\pi}2+k\pi}\frac{\cos u}{\sqrt[4]{u}}\mathrm du$$

then $(a_k)$ is an alternating sequence (by the change of sign of the cosine) and $(|a_k|)\to 0$ (because $\sqrt[4]{u}$ is unbounded), hence the series $\sum_{k=0}^\infty a_k$ converges by the alternating series test.

Now observe that

$$0\le\int_0^{\pi/2}\frac{\cos u}{\sqrt[4]{u}}\mathrm du=\int_0^{\pi^2/4} \sqrt t\cos (t^2)\mathrm dt<\infty$$

Hence

$$\int_0^\infty\sqrt t\cos(t^2)\mathrm dt=\int_0^{\pi^2/4} \sqrt t\cos (t^2)\mathrm dt+\sum_{k=0}^\infty a_k<\infty$$

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  • $\begingroup$ Try to integrate by parts, recalling that $\cos u$ has bounded integrals. $\endgroup$ – Siminore Apr 6 '17 at 8:32
  • $\begingroup$ How is $a_k$ alternating if $\cos u$ is always negative on the intervals $(\pi/2+2\pi k, 3\pi/2+2\pi k)$? $\endgroup$ – uniquesolution Apr 6 '17 at 8:36
  • $\begingroup$ @uniquesolution I had a typo, fixed. $\endgroup$ – Masacroso Apr 6 '17 at 8:38
  • $\begingroup$ You can use Abel's criterion for integrals. $\endgroup$ – Guy Fsone Apr 6 '17 at 8:51
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An alternative way. Note that $$I=\frac{1}{2}\int_{0}^{\infty}u^{-1/4}\cos\left(u\right)du=\frac{1}{2}\textrm{Re}\left(\int_{0}^{\infty}u^{-1/4}e^{-iu}du\right)$$ then taking $z=iu$ we get $$I=\frac{1}{2}\textrm{Re}\left(i^{-3/4}\int_{0}^{\infty}z^{-1/4}e^{-z}du\right)=\frac{1}{2}\textrm{Re}\left(i^{-3/4}\Gamma\left(\frac{3}{4}\right)\right).$$ It is important to observe that we may continue to integrate from $0$ to $\infty$, due to the analytic continuation. Since $$i^{-3/4}=\sin\left(\frac{\pi}{8}\right)-i\cos\left(\frac{\pi}{8}\right)$$ we get $$I=\color{red}{\frac{1}{2}\sin\left(\frac{\pi}{8}\right)\Gamma\left(\frac{3}{4}\right)}$$ so the integral converges.

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You are on the right way. Start from $$ \int_0^{+\infty} \frac{\cos u}{\sqrt[4]{u}}du $$ and apply Dirichlet's test (see page 34 of this file). The function $u \mapsto 1/\sqrt[4]{u}$ is bounded and monotonically decaying to zero at infinity. Here you have a little issue at $u=0$, but this is easily analyzed directly: so you can integrate over $[1,+\infty)$.

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  • $\begingroup$ oh, right... I forget the zero, lol. $\endgroup$ – Masacroso Apr 6 '17 at 8:40

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