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We only consider real functions here. It can be shown that \begin{eqnarray*} \int_{0}^{\infty}dx\frac{\sin x}{\sqrt{x}} & = & \int_{0}^{\infty}dx\frac{\cos x}{\sqrt{x}}=\sqrt{\frac{\pi}{2}} \end{eqnarray*}

It is obvious that the integration $\int_{0}^{\infty}dx\sqrt{x}\cos x$ is not convergent. But here I will show another way to calculate $\int_{0}^{\infty}dx\sqrt{x}\cos x$. We can see \begin{eqnarray*} \int_{0}^{\infty}dx\sqrt{x}\cos(kx) & = & \int_{0}^{\infty}dx\frac{1}{\sqrt{x}}\frac{\partial}{\partial k}\sin(kx)=\int_{0}^{\infty}dx\frac{1}{\sqrt{x}}\lim_{\Delta k\to0}\frac{\sin[(k+\Delta k)x]-\sin(kx)}{\Delta k} \end{eqnarray*}

Suppose that the integration process and the limit process can be exchanged, then we have \begin{eqnarray*} \int_{0}^{\infty}dx\sqrt{x}\cos(kx) & = & \lim_{\Delta k\to0}\int_{0}^{\infty}dx\frac{1}{\sqrt{x}}\frac{\sin[(k+\Delta k)x]-\sin(kx)}{\Delta k}\\ & = & \lim_{\Delta k\to0}\frac{1}{\Delta k}\bigg(\int_{0}^{\infty}dx\frac{\sin[(k+\Delta k)x]}{\sqrt{x}}-\int_{0}^{\infty}dx\frac{\sin(kx)}{\sqrt{x}}\bigg)\\ & = & \lim_{\Delta k\to0}\frac{1}{\Delta k}\bigg(\frac{1}{\sqrt{k+\Delta k}}-\frac{1}{\sqrt{k}}\bigg)\sqrt{\frac{\pi}{2}}\\ & = & \sqrt{\frac{\pi}{2}}\frac{d}{dk}\frac{1}{\sqrt{k}}\\ & = & -\frac{1}{2}\sqrt{\frac{\pi}{2}}\frac{1}{k\sqrt{k}} \end{eqnarray*}

i.e. \begin{eqnarray*} \int_{0}^{\infty}dx\sqrt{x}\cos(kx) & = & -\frac{1}{2}\sqrt{\frac{\pi}{2}}\frac{1}{k\sqrt{k}} \end{eqnarray*}

then we have \begin{eqnarray*} \int_{0}^{\infty}dx\sqrt{x}\cos x & = & \bigg(\int_{0}^{\infty}dx\sqrt{x}\cos(kx)\bigg)_{k=1}=-\frac{1}{2}\sqrt{\frac{\pi}{2}} \end{eqnarray*}

which conflicts with the non-convergence of $\int_{0}^{\infty}dx\sqrt{x}\cos x$. This confliction may be due to the fact that the integration process and the limit process can not be exchanged. So my question (which may be naive to mathematicians) is: what is the necessary and sufficient condition for exchanging the integration process and the limit process?

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    $\begingroup$ I'm not aware of a necessary and sufficient condition for exchanging the limit and the integral. But there is many theorems that gives sufficient conditions $\endgroup$
    – Tryss
    Jul 14, 2016 at 16:06
  • $\begingroup$ Please: write \cos x, not \mathrm{cos}x. Don't you see the conspicuous lack of proper spacing with the latter? Contrast 2\cos x with 2\mathrm{cos}x: $$ 2\cos x \quad\text{versus}\quad2\mathrm{cos}x $$ The spacing is different in 2\cos(x): $$2\cos(x)$$ \mathrm{} doesn't have any context-depended spacing conventions built in to it. I edited the question. $\qquad$ $\endgroup$ Jul 14, 2016 at 16:22
  • $\begingroup$ @MichaelHardy...Thanks for your suggestions! I have replaced \mathrm{cos}x by \cos x everywhere. $\endgroup$ Jul 15, 2016 at 3:05
  • $\begingroup$ @Tryss...Can you recommend some mathematical book that involves sufficient condition for exchanging the limit and the integral? Thank you very much. $\endgroup$ Jul 15, 2016 at 3:24
  • $\begingroup$ Any book on real analysis? en.wikipedia.org/wiki/Dominated_convergence_theorem for just one such theorem. $\endgroup$ Jul 15, 2016 at 3:43

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