Integration – Definite Integral of $\cos (x)/ \sqrt{x}$

Question

How do I integrate
$$ \int^{+\infty}_0 \frac{\cos(x)}{\sqrt{x}} dx$$
I tried setting $u = x^{-1/2}$ and $dv = \cos(x)dx$. Then I integrate by part twice to get:
$$ \int^\infty_0 \frac{\cos x}{\sqrt{x}} dx = \lim_{x \rightarrow \infty} \left[\sqrt{x} \sin(x) – \frac{1}{2} \sqrt{x} \cos(x)\right] – \frac{1}{4} \int^\infty_0 \frac{\cos x}{\sqrt{x}} dx $$
Hence:
$$ \frac{5}{4} \int^\infty_0 \frac{\cos x}{\sqrt{x}} dx = \lim_{x \rightarrow \infty} \left[\sqrt{x} \sin(x) – \frac{1}{2} \sqrt{x} \cos(x)\right] $$
I'm not sure how to evaluate the terms on the right. Also, if you think I made a mistake above, please help me point it out.

Answer 1

Setting $\sqrt{x}=t$, we obtain $$I = \int_0^{\infty} \dfrac{\cos(t^2)}{t} (2tdt) = 2\int_0^{\infty} \cos(x^2)dx = 2 \text{Real}\left(\int_0^{\infty} e^{ix^2}dx\right)$$ Consider the integral $$F_R = \int_0^{R} e^{iz^2}dz + \int_{C_R} e^{iz^2}dz + \int_{Re^{i\pi/4}}^0 e^{iz^2}dz = \int_0^{R} e^{iz^2}dz + \int_{C_R} e^{iz^2}dz + \int_{R}^0 e^{i\left(te^{i\pi/4}\right)^2} e^{i\pi/4}dt $$ where $C_R$ is one-eighth of the circle in the first quadrant. We have $F_R = 0$, since $e^{iz^2}$ is analytic inside $[0,R]\cup C_R \cup[iR,0]$. Further, we have $$\left \vert\int_{C_R} e^{iz^2}dz \right \vert \leq \int_{0}^{\pi/4} e^{-R^2\sin(2\theta)} R d\theta$$ Hence, $$\lim_{R \to \infty}\left \vert\int_{C_R} e^{iz^2}dz \right \vert \leq \lim_{R \to \infty}\int_{0}^{\pi/4} e^{-R^2\sin(2\theta)} R d\theta \leq \int_{0}^{\pi/4} \lim_{R \to \infty}e^{-R^2\sin(2\theta)} R d\theta = 0$$ We hence have $$0 = \lim_{R \to \infty}F_R = \int_0^{\infty} e^{iz^2}dz + e^{i\pi/4} \int_{\infty}^0 e^{-t^2}dt = \int_0^{\infty} e^{iz^2}dz - \dfrac{\sqrt{\pi}}2e^{i \pi /4}$$ This gives us $$\int_0^{\infty} e^{iz^2}dz = \dfrac{\sqrt{\pi}}2 e^{i\pi/4}$$ Hence, our integral is $$I = 2 \times \dfrac{\sqrt{\pi}}2 \times \cos\left(\dfrac{\pi}4\right) = \sqrt{\dfrac{\pi}{2}}$$