Exercise 6.13

Question

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Exercise 13: Define $$f(x)=\int_x^{x+1}\sin t^2\,dt.$$

(a) Prove that $\big|f(x)\big|<1/x$ if $x>0$.

(b) Prove that $$2xf(x)=\cos x^2-\cos(x+1)^2+r(x)$$ where $\big|r(x)\big|<c/x$ and $c$ is a constant.

(c) Find the upper and lower limits of $xf(x)$, as $x\rightarrow\infty$.

(d) Does $\int_0^{\infty}\sin t^2\,dt$ converge?

analambanomenos · Accepted Answer

(a) Following the hint (substituting $u$ for $t^2$, integration by parts using $\sin u$ and $1/\bigl(2\sqrt{u}\bigr)$, replacing $|\cos u|$ with 1), we get
    \begin{align*}
        f(x) &= \int_{x^2}^{(x+1)^2}\frac{\sin u}{2\sqrt{u}}\,du \
        &= -\frac{\cos(x+1)^2}{2(x+1)}+\frac{\cos x^2}{2x}-\int_{x^2}^{(x+1)^2}\frac{\cos u}{4u^{3/2}}\,du \
        \big|f(x)\big| &\le \frac{\big|\cos(x+1)^2\big|}{2(x+1)}+\frac{|\cos x^2|}{2x}+\int_{x^2}^{(x+1)^2}\frac{|\cos u|}{4u^{3/2}}\,du \
        &< \frac{1}{2(x+1)}+\frac{1}{2x}+\int_{x^2}^{(x+1)^2}\frac{1}{4u^{3/2}}\,du \
        &= \frac{1}{2(x+1)}+\frac{1}{2x}+\bigg(\frac{1}{2x}-\frac{1}{2(x+1)}\bigg) \
        &= \frac{1}{x}
    \end{align*}
    (The strict inequality is due to the fact that $|\cos u|$ will not be constantly equal to its maximum possible value 1 throughout the interval of integration.)

(b) From the results in part (a), we get
    \begin{align*}
        2xf(x) &< \cos x^2 -\frac{x\cos(x+1)^2}{x+1}+\frac{1}{x+1} \
        r(x)=2xf(x)-\cos x^2+\cos(x+1)^2 &< \frac{\cos(x+1)^2}{x+1}+\frac{1}{x+1} \
    \big|r(x)\big| &< \frac{2}{x+1} < \frac{2}{x}
    \end{align*}

(c) (partial) From part (b), we get that as $xightarrow\infty$, $$xf(x)\approx\frac{\cos x^2-\cos(x+1)^2}{2}.$$ Hence $xf(x)$ lies (more or less) between $(-1-1)/2=-1$ and $(1+1)/2=1$.

(d) Note that $\sin t^2$ is positive between $\sqrt{n\pi}$ and $\sqrt{(n+1)\pi}$ for any even integer $n$ and negative for any odd integer $n$. Hence $\int_0^{\infty}\sin t^2\,dt$ can be 
    reduced to an alternating series, whose terms satisfy $$\Bigg|\int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}}\sin t^2\,dt\Bigg|<\sqrt{n\pi}-\sqrt{(n+1)\pi}=\frac{\sqrt{\pi}}{\sqrt{n}+\sqrt{n+1}}$$
    which goes to 0 as $nightarrow\infty$. Hence, by Theorem 3.43, the integral converges.

Exercise 6.13

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Exercise 6.13

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