$\int^\infty_0 \frac{\cos(x)}{\sqrt{x}}\,dx$ Evaluate using Fresnel Integrals

Question

$\int^\infty_0 \frac{\cos(x)}{\sqrt{x}}\,dx$ Evaluate using Fresnel Integrals

(For reference the $\cos$ Fresnel integral is $\int^\infty_0 \cos(x^2)\, dx = \frac{\sqrt{2 \pi}}{4}$)

I've tried integration by parts but just ended up getting $-x\cos(x)$ for my final integration which doesn't help.

I suppose we want to some how get $\cos(u^2)$ into the integrand, but I'm stupid and can't figure out how.

Mathematica says the answer is $\frac{\sqrt{2\pi}}{2}$

Any help would be appreciated!

Answer 1

Using Fresnel Integrals

Substituting $x=u^2$, we get $$ \int_0^\infty\frac{\cos(x)}{\sqrt{x}}\mathrm{d}x =2\int_0^\infty\cos(u^2)\,\mathrm{d}u $$ As shown in this answer, $$ \int_0^\infty\cos(u^2)\,\mathrm{d}u=\sqrt{\frac\pi8} $$ Therefore, $$ \int_0^\infty\frac{\cos(x)}{\sqrt{x}}\mathrm{d}x=\sqrt{\frac\pi2} $$

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Alternate Approach

As a check, we can use contour integration to show that since $\frac{e^{iz}}{\sqrt{z}}$ has no singularities in the plane minus the negative real axis, we have $$ \begin{align} \int_0^\infty\frac{\cos(x)}{\sqrt{x}}\mathrm{d}x &=\mathrm{Re}\left(\int_0^\infty\frac{e^{ix}}{\sqrt{x}}\mathrm{d}x\right)\\ &=\mathrm{Re}\left(\frac{1+i}{\sqrt2}\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\mathrm{d}x\right)\\ &=\frac1{\sqrt2}\Gamma\left(\frac12\right)\\ &=\sqrt{\frac\pi2} \end{align} $$

Answer 2

Let $x=u^2$ and substitute. It's straightforward once you know the Fresnel integral you mentioned.

Answer 3

$$\int_0^{\infty} \frac{\cos x}{\sqrt{x}}dx=\int_0^{\infty} \frac{cos t^2}{t}dt^2$$