# What am I doing wrong here ? $\int_0^{\infty}\sin(x^2)dx$

What am I doing wrong here ? $\displaystyle\int_0^{\infty}\sin(x^2)dx$

I made the following: $\displaystyle\int_0^1\sin(x^2)dx<+\infty$ then I take $b>1$ and:

$\bigg|\displaystyle\int_1^b\sin(x^2)dx\bigg|\overset{u=x^2}=\frac12\bigg|\int_{1^b}^{b^2}\frac{\sin(u)}{\sqrt{u}}du\bigg|\overset{IBP}=\frac12\bigg|\frac{-\cos(u)}{\sqrt{u}}+\int_{1^2}^{b^2}\cos(u)d(u^{-1/2})\bigg|\le\frac12\bigg|\frac1b+\frac11+\int_{1^2}^{b^2}d(u^{-1/2})\bigg|=\frac1b$

So, if $b\to\infty$ then $\bigg|\displaystyle\int_1^b\sin(x^2)dx\bigg|\to0$

Hence;

$\displaystyle\int_0^{\infty}\sin(x^2)dx=\int_0^1\sin(x^2)dx$

but this is not true. (One can also take an arbitrary value for example $a>0$, instead of $1$ for the lower limit, this would also lead to a contradiction.)

The inequality

$$\frac12\bigg|\frac{-\cos(u)}{\sqrt{u}}+\int_{1^2}^{b^2}\cos(u)d(u^{-1/2})\bigg|\le\frac12\bigg|\frac1b+\frac11+\int_{1^2}^{b^2}d(u^{-1/2})\bigg|=\frac 1b$$

is incorrect. You can try some big number in Wolfram alpha to see that for example

$$\int_1^{1000} \sin(x^2) dx \sim 0.31592 > \frac 1{1000}.$$

What you had done wrong is that you approximate under the $|\cdot |$ sign. You should only get

$$\frac12\bigg|\frac{-\cos(u)}{\sqrt{u}}+\int_{1^2}^{b^2}\cos(u)d(u^{-1/2})\bigg| \leq \frac 12 \bigg(\frac 1b + 1\bigg) + \bigg| \frac 12 \int_1^{b^2} d (u^{-1/2}) \bigg|$$ $$\leq \frac 12 \bigg( \frac 1b + 1 + \big| \frac 1b -1\big|\bigg)$$

You cannot cancel that 1 from here.

• Hmm, this would only show then that the integral exist, right ? – OBDA Jun 13 '14 at 11:28
• @OBDA: Not really, this only show that the integral is bounded for all $b$. Not sure if limit exist as $b\to \infty$. – user99914 Jun 13 '14 at 12:29
• but if your last inequality is correct then for large $b$: $|\frac1b-1|=1-\frac1b$ and hence the integral from $1$ to $\infty$ is bounded by $\frac12(\frac1b+1+1-\frac1b)=1$, or not ? – OBDA Jun 13 '14 at 12:41
• @OBDA: Yes, it's bounded. But you still can't say that it has a limit. Just like $\sin x$, it's bounded between $-1$ and $1$, but $\lim_{x\to \infty} \sin x$ does not exist. – user99914 Jun 13 '14 at 12:49

The error in your calculation is in the inequality.

I think you thought of the triangle inequality and maximized the cosine, but that will not work - it would lead you to see that $b\mapsto \int_1^b \sin(x^2)dx$ is bounded which is true for $\int_1^b \sin(x)dx$ too.