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.

  • $\begingroup$ Breaking up a limit into a sum of two limits is valid only when both of the limits in the sum exist; $\sqrt{x}(\sin x - \frac12 \cos x)$ doesn't converge as $x\to\infty$. That is to say, this integration by parts can't be carried out. $\endgroup$ – Eugene Shvarts May 13 '15 at 0:13

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 $$I = 2 \times \dfrac{\sqrt{\pi}}2 \times \cos\left(\dfrac{\pi}4\right) = \sqrt{\dfrac{\pi}{2}}$$

  • $\begingroup$ The interchange of the limit and integral requires justification. An straightforward way is to note that on $[0,\pi/4]$, we have $\sin 2x\ge 4x/pi$. Thus, $\int_0^{\pi/4}e^{-R^2\sin 2x}Rdx \le \int_0^{\pi/4}e^{-R^2 4x/\pi}Rdx=\frac{\pi}{4}\frac{1-e^{-R^2}}{R}\to 0$ as $R\to \infty$. Nice solution otherwise! $\endgroup$ – Mark Viola Jul 7 '15 at 5:31

It doesn't look to me like you did integration by parts right. Doing it once, I get

$$u = x^{-1/2}\qquad v = \sin(x)$$ $$du = -\frac{1}{2}x^{-3/2}\;dx \qquad dv =\cos(x)\;dx$$

so $$\int_0^\infty \frac{\cos(x)}{\sqrt{x}}\;dx = \left[\frac{\sin(x)}{\sqrt{x}}\right]_0^\infty +\frac{1}{2} \int_0^\infty \frac{\sin(x)}{x^{3/2}}\;dx$$

which is not any easier to evaluate.

One way to do the integration is to substitute $u=\sqrt{x}$, so $x=u^2$ and $du = \frac{1}{2\sqrt{x}}\;dx$, so $$\int_0^\infty \frac{\cos(x)}{\sqrt{x}}\;dx = 2 \int_0^\infty \cos(u^2)\;du$$

The left hand side is twice the limit of the Fresnel Integral $C(t)$ as $t\to\infty$, so

$$2 \int_0^\infty \cos(u^2)\;du = 2 \sqrt{\frac{\pi}{8}} = \sqrt{\frac{\pi}{2}}$$

  • $\begingroup$ Bravo to your suggestion of the Fresnel Integral. I went that way with this integral once, but I didn't know how to integrate that integral. Thank you for the enlightenment. $\endgroup$ – Huy Nguyen May 13 '15 at 0:26
  • $\begingroup$ @HuyNguyen Look at my answer for evaluating the Fresnel integral, the right way, $\endgroup$ – Leg May 13 '15 at 0:42

One way to evaluate the integral would be to note that it is the real part of $$ \int^{+\infty}_0 \frac{e^{ix}}{\sqrt{x}} dx$$ and use the gamma function to evaluate this.

  • $\begingroup$ I actually reduced my integral from that form. I was doing a Fourier transform integral. $\endgroup$ – Huy Nguyen May 13 '15 at 0:24
  • $\begingroup$ @HuyNguyen Well the above can be evaluated using the gamma function (substituting $x = iu$ and deforming contours to get to the gamma function form). $\endgroup$ – David Foster May 13 '15 at 0:27
  • $\begingroup$ Can you put your suggestion into an answer? I'm not very familiar with deforming contours. It will be very helpful for related problems. $\endgroup$ – Huy Nguyen May 13 '15 at 0:32

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.