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!


3 Answers 3


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

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  • $\begingroup$ Although one can reason easily that $A_n$ decreases, is there a way to show it formally? $\endgroup$
    – MT_
    May 14, 2014 at 12:19
  • 1
    $\begingroup$ @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. $\endgroup$ May 14, 2014 at 12:25
  • $\begingroup$ @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. $\endgroup$ May 14, 2014 at 12:50
  • $\begingroup$ @MarkMcClure You're right, it's a bit harder than I thought. $\endgroup$ 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.

  • $\begingroup$ 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... $\endgroup$
    – user475680
    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.

  • $\begingroup$ Does this work? $\endgroup$
    – MT_
    May 14, 2014 at 12:17
  • $\begingroup$ 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 $t=x^2$). $\endgroup$
    – user37238
    May 14, 2014 at 12:23
  • $\begingroup$ @MichaelT Why don't you try it? :P $\endgroup$ May 14, 2014 at 12:24
  • $\begingroup$ @Goos It was an early morning browse, I had no time at the time. $\endgroup$
    – MT_
    May 14, 2014 at 21:34

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