# Does $\int _1^{\infty }\left(\sin \left(x^2\right)\right)dx$ converge or diverge? [duplicate]

I'm in need of some assistance regarding a question in my Calculus textboox:

Find if the following converges or diverges without calculating the integral:

$$\int _1^{\infty }\left(\sin \left(x^2\right)\right)dx$$

I tried using several methods, including the convergence test but with no luck.

Any help is appreciated, Thx!

• Not an answer, but background info: Your integral is related to limit of the Fresnel integral $S(x).$ May 14, 2014 at 12:19
• May 14, 2014 at 22:12

Yet another hint: Think in terms of alternating series:
$$\int_0^{\infty} \sin(x^2) \, dx = \sum_{n=1}^{\infty} A_n.$$

• Although one can reason easily that $A_n$ decreases, is there a way to show it formally?
– MT_
May 14, 2014 at 12:19
• @MichaelT Yes. The zeros are where $x^2 = 2\pi k$. Thus the $A_n$ are bounded by something like $\sqrt{2 \pi (k+1)} - \sqrt{2 \pi k}$ in absolute value. May 14, 2014 at 12:25
• @Goos I suppose you mean $k\pi$, not $2k\pi$. That shows that $A_k\rightarrow 0$, but MichaelT asks why are they decreasing? Perhaps a bit trickier. May 14, 2014 at 12:50
• @MarkMcClure You're right, it's a bit harder than I thought. May 14, 2014 at 12:56

Note that the integral is equal to

$$\operatorname{Im}{\left [\int_1^{\infty} dx \, e^{i x^2} \right ]}$$

Because the integral over $[0,1]$ is in fact finite, the question is equivalent to considering whether the following integral converges:

$$\operatorname{Im}{\left [\int_0^{\infty} dx \, e^{i x^2} \right ]}$$

To show that it indeed converges, consider the following contour integral in the complex plane:

$$\oint_C dz \, e^{i z^2}$$

where $C$ is a $45$-degree circular sector in the upper right quadrant, along the real axis, of radius $R$. The contour integral is then equal to

$$\int_0^R dx \, e^{i x^2} + i R \int_0^{\pi/4} d\theta \, e^{i \theta} \, e^{i R^2 e^{i 2 \theta}} + e^{i \pi/4} \int_R^0 dt \, e^{-t^2}$$

As $R \to \infty$, we can show that the second integral has a magnitude bounded by

$$R \int_0^{\pi/4} d\theta \, e^{-R^2 \sin{2 \theta}} \le \frac{R}{2} \int_0^{\pi/2} d\theta \, e^{-2 R^2 \theta/\pi} \le \frac{\pi}{4 R}$$

Thus the second integral vanishes as $R\to\infty$. By Cauchy's theorem, the contour integral is zero; therefore

$$\int_0^{\infty} dx \, e^{i x^2} = e^{i \pi/4} \int_0^{\infty} dt \, e^{-t^2}$$

which converges. Thus, the original integral converges.

• This looks like a cool way to solve it, unfortunately there are some things here that I haven't learned yet so I can't use it... May 14, 2014 at 12:12

Hint : Write $\int_1^M \sin(x^2)\text{d}x$ as $\int_1^M \frac{2x}{2x}\sin(x^2)\text{d}x$ and then use integration by parts.

• Does this work?
– MT_
May 14, 2014 at 12:17
• Yes. It is exactly the same idea in the answer of Davide Giraudo in math.stackexchange.com/questions/105107/… (I just didn't use the change of variable $t=x^2$). May 14, 2014 at 12:23
• @MichaelT Why don't you try it? :P May 14, 2014 at 12:24
• @Goos It was an early morning browse, I had no time at the time.
– MT_
May 14, 2014 at 21:34