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$\int^\infty_0 \frac{\cos(x)}{\sqrt{x}}\,dx$ 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!

Daniel Fischer
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chriskinda
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3 Answers3

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

Angelo
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  • Then you get $\frac{cos(t^2)}{t}dt^2$ as noted below which doesn't converge as far as I can tell. – chriskinda Feb 01 '15 at 12:11
  • The way I see it is: Let $x=t^2$, say (instead of $u$). Then $dx=2t, dt$ Now when $x=0$, $t=0$, and when $x=\infty$, $t=\infty$. So the new limits in the $t$-integral are the same as the old ones. Next $$\int_{0}^\infty \frac{\cos x}{\sqrt{x}}, dx=\int_{0}^\infty \frac{\cos t^2}{t},(2t, dt) = 2\int_{0}^\infty \cos{t^2}, dt.$$ So, my point is that IF you know about the Fresnel integral then you are done, but if you want to derive the Fresnel integral formula then that is another question. Maybe I misunderstood your question!? – Angelo Feb 01 '15 at 12:28
  • Nope! You got it 100%, I'm just feeling especially sleepy today. ;) Thanks you! – chriskinda Feb 01 '15 at 12:45
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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, $$ \int_0^\infty\frac{\cos(x)}{\sqrt{x}}\mathrm{d}x=\sqrt{\frac\pi2} $$

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 $$ \begin{align} \int_0^\infty\frac{\cos(x)}{\sqrt{x}}\mathrm{d}x &=\mathrm{Re}\left(\int_0^\infty\frac{e^{ix}}{\sqrt{x}}\mathrm{d}x\right)\\ &=\mathrm{Re}\left(\frac{1+i}{\sqrt2}\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\mathrm{d}x\right)\\ &=\frac1{\sqrt2}\Gamma\left(\frac12\right)\\ &=\sqrt{\frac\pi2} \end{align} $$

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

Road Human
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  • If there's really $;dt^2;$ there then both integrals are exactly the same, both their value and, as far as I can see, the difficulty to evaluate them. – Timbuc Feb 01 '15 at 11:54
  • you forgot to see that $dx= 2u, du$ so the u's cancel leaving you with twice the Fresnel integral. – Angelo Feb 01 '15 at 12:05
  • If you use the substitution $x = t^2,$ then you must replace $dx$ with $2t,dt$ to get $2\int \cos(t^2) ,dt$. Please correct, or delete, your answer. – amWhy Feb 01 '15 at 12:12
  • @Timbuc Maybe the problem is symbolic!? $dt^2$ as you have it means $d(t^2)$ which is $2t, dt$, if you want to think of this as differentials (but you shouldn't). Anyhow the point is you don't get $dt^2$ but $2t, dt$. – Angelo Feb 01 '15 at 12:37
  • @Angelo As I see this answer is precisely as a symbolic substitution, and certainly not a mathematical one: Road is simply substituting symbolically $;x\to t^2;$ , which yields a correct integral on the right (under this assumption!) but just as difficult to solve as the original one. Of course, substituting as in integral calculus we get at once $;x=u^2\implies dx =2udu;$ , as you mention. – Timbuc Feb 01 '15 at 12:40
  • so are you more interested in the Fresnel integral evaluation or did you assume that? – Angelo Feb 01 '15 at 13:42