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When doing an exercise, I found that a sequence $(u_n)$ satisfies the following $$ u_n \underset{n\to + \infty}{\sim} \frac{1}{n^{\alpha/2}} \int_0^{n^\alpha} \frac{\cos(x)}{\sqrt{x}} \text{d}x, $$ for $\alpha \in [1,2]$ (the original question is to determine if $\sum u_n$ is convergent or not). I know that \begin{align} \int_0^{+\infty} \frac{\cos(x)}{\sqrt{x}} \text{d}x = \sqrt{\frac{\pi}{2}}, \tag{1} \end{align} so that $$ u_n \underset{n\to + \infty}{\sim} \frac{1}{n^{\alpha/2}} \sqrt{\frac{\pi}{2}}, $$ from which it is easy to conclude. I don't think that the purpose of the exercise is to derive $(1)$ (see this MSE question or that one), one only needs to prove that $\int_0^{+\infty} \frac{\cos(x)}{\sqrt{x}} \text{d}x$ exists (which is easy thanks to integration by parts) and not equal to $0$. So my question is, how to derive (in a simple and/or short manner) that $$ \int_0^{+\infty} \frac{\cos(x)}{\sqrt{x}} \text{d}x \neq 0.$$

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user37238
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The original integral is convergent (as an improper Riemann integral) by Dirichlet's criterion, since $\cos x$ has a bounded primitive while $\frac{1}{\sqrt{x}}$ is decreasing to zero. It follows that: $$I=\int_{0}^{+\infty}\frac{\cos(x)}{\sqrt{x}}\,dx \stackrel{IBP}{=}\frac{1}{2}\int_{0}^{+\infty}\frac{\sin x}{x^{3/2}}\,dx = \frac{1}{2}\int_{0}^{\pi}\sin(x)\sum_{n\geq 0}\frac{(-1)^n}{(x+n\pi)^{3/2}}\,dx$$ but the last sum is a positive function on $(0,\pi)$, since: $$\sum_{n\geq 0}\frac{(-1)^n}{(x+n\pi)^{3/2}}=\sum_{n\geq 0}\left(\frac{1}{(x+2n\pi)^{3/2}}-\frac{1}{(x+(2n+1)\pi)^{3/2}}\right).$$ It follows that $I>0$.

Another chance is given by the Laplace transform.

Since $\mathcal{L}(\cos x)=\frac{s}{1+s^2}$ and $\mathcal{L}^{-1}\left(\frac{1}{\sqrt{x}}\right)=\frac{1}{\sqrt{\pi s}}$,

$$ I = \frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{\sqrt{s}}{1+s^2}\,ds $$ and the last integral is trivially positive.

Jack D'Aurizio
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  • I was about to write the same answer, yet with the intermediate step: $^\sum_{k\ge 0}\int_{k\pi}^{(k+1)\pi}$ and then collecting the terms pairwise, exaclty as you did – b00n heT May 03 '16 at 12:59
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    Or to paraphrase Norbert Wiener, the improper integral is an alternating sum of positive decreasing terms, hence positive. – Umberto P. May 03 '16 at 13:02
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    You don't need to do the integration to see that this is positive, namely $\int_{2k\pi}^{2(k+1)\pi}\frac{\cos(x)}{\sqrt{x}}dx$ is always positive because the integrand in $\int_{2k\pi}^{(2k+1)\pi}\frac{\cos(x)}{\sqrt{x}}dx$ is pointwise greater (in absolute value) than the integrand in $\int_{(2k+1)\pi}^{(2k+2)\pi}\frac{\cos(x)}{\sqrt{x}}dx$. – Michael Burr May 03 '16 at 13:05