Prove that


I have some question. Using this, find this integral is not converge, I'm wrong?

Thank you everyone


First make the substitution $x=u^2$ to get:

$\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x} \right) {dx}=\int _{0}^{\infty }\!2\,\sin \left( {u}^{2} \right) \sin \left( u \right) u{du}$,

$\displaystyle=-\int _{0}^{\infty }\!u\cos \left( u \left( u+1 \right) \right) {du}+ \int _{0}^{\infty }\!u\cos \left( u \left( u-1 \right) \right) {du}$,

and changing variable again in the second integral on the R.H.S such that $u\rightarrow u+1$ this becomes:

$=\displaystyle\int _{0}^{\infty }\!-u\cos \left( u \left( u+1 \right) \right) {du}+ \int _{-1}^{\infty }\!\left(u+1\right)\cos \left( u \left( u+1 \right) \right) {du} $,

$\displaystyle=\int _{0}^{\infty }\!\cos \left( u \left( u+1 \right) \right) {du}+ \int _{-1}^{0}\! \left( u+1 \right) \cos \left( u \left( u+1 \right) \right) {du} $.

Now we write $u=v-1/2$ and this becomes:

$\displaystyle\int _{1/2}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}+\int _{-1/ 2}^{1/2}\! \left( v+1/2 \right) \cos \left( {v}^{2}-1/4 \right) {dv}=$

$\displaystyle \left\{\int _{0}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}\right\}$

$\displaystyle +\left\{\int _{-1/2} ^{1/2}\!v\cos \left( {v}^{2}-1/4 \right) {dv}+\int _{-1/2}^{0}\!1/2\, \cos \left( {v}^{2}-1/4 \right) {dv}-1/2\,\int _{0}^{1/2}\!\cos \left( {v}^{2}-1/4 \right) {dv}\right\},$

but the second curly bracket is zero by symmetry and so:

$\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x} \right) {dx}=\displaystyle \int _{0}^{\infty }\!\cos \left( {v}^{2}-1/4 \right) {dv}$,

$\displaystyle =\int _{0}^{ \infty }\!\cos \left( {v}^{2} \right) {dv}\cos \left( 1/4 \right) + \int _{0}^{\infty }\!\sin \left( {v}^{2} \right) {dv}\sin \left( 1/4 \right) $.

We now quote the limit of Fresnel integrals:

$\displaystyle\int _{0}^{\infty }\!\cos \left( {v}^{2} \right) {dv}=\int _{0}^{ \infty }\!\sin \left( {v}^{2} \right) {dv}=\dfrac{\sqrt{2\pi}}{4}$,

to obtain:

$\displaystyle \int _{0}^{\infty }\!\sin \left( x \right) \sin \left( \sqrt {x} \right) {dx}=\dfrac{\sqrt{2\pi}}{4}\left(\cos\left(\dfrac{1}{4}\right)+\sin\left(\dfrac{1}{4}\right)\right)=\dfrac{\sqrt{\pi}}{2}\sin{\left(\dfrac{3\pi-1}{4}\right)}$.

  • $\begingroup$ When you replace one of $u$ by $u+1$, you changed the way how $u$ goes to infinity, so this result does not conflict my assertion. $\endgroup$ – Ma Ming May 21 '13 at 13:32
  • $\begingroup$ Could you be more specific, I let $u\rightarrow u+1$ in only the right integral in the third line simply because it is a dummy variable; that is, it is unrelated to the other integral. How has the way it goes to infinity changed? Why does that effect the result? $\endgroup$ – Graham Hesketh May 21 '13 at 13:39
  • $\begingroup$ Let $F(u)=A(u)+B(u)=\int_0^u (a+b)$, when you replace $u$ by $u+1$ in $b$, what basically you obtain is $F_1(u)=A(u)+B(u+1)$. If $A$ and $B$ both converges then this doesn't matter. $\endgroup$ – Ma Ming May 21 '13 at 13:45
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    $\begingroup$ @GrahamHesketh What you need to verify is $A(u)$ and $B(u)$ both converge (at least one), then you will get $\lim_{u\to\infty}F(u)=\lim_{u\to\infty}A(u)+\lim_{u\to\infty}B(u)=\lim_{u\to \infty} A(u)+\lim_{u\to\infty}B(u+1)$. Otherwise, I suppose one can produce any value as the result. $\endgroup$ – Ma Ming May 22 '13 at 9:18
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    $\begingroup$ For example, $$\begin{array}{ll} 0 & = \int_0^\infty 0du \\ & =\int_0^\infty 1du+\int_0^\infty -1du \\ & = \int_{-1}^\infty1du+\int_0^\infty -1du \\ & = \int_{-1}^01du+\int_0^\infty(1-1)du \\ & = 1\end{array}$$ hence $0=1$, would be the result of a similar computation. $\endgroup$ – anon May 24 '13 at 1:30

Numerical calculation shows that the graph of

$$ y = \int_{0}^{x} \sin t \sin \sqrt{t} \, dt $$

is given by

enter image description here

Though a graph cannot constitute a proof, it strongly suggests that $y$ cannot converge as $x \to \infty$. Indeed, we can show that

$$ y(x) = -\cos x \sin \sqrt{x} + \frac{\sqrt{\pi}}{2} \sin \left( \frac{3\pi - 1}{4} \right) + o(1) $$

as $x \to \infty$. Thus in ordinary sense, the integral

$$ \int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx $$

diverges. But if we understand the integral in Abel summation senas as follows

$$ \int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx := \lim_{s \to 0^{+}} \int_{(0, \infty)} \sin x \sin \sqrt{x} \; e^{-sx} \, dx, $$

then the oscillating part vanishes and we obtain the identity. In fact, it converges to the proposed limit as weak as in Cesaro summation sense.

Now let us show that the integral converges in Able summation sense. By some calculation, we have

\begin{align*} \int_{0}^{\infty} \sin x \sin \sqrt{x} \; e^{-sx} \, dx &= \int_{0}^{\infty} 2x \sin x \sin (x^2) \; e^{-sx^{2}} \, dx \\ &= \int_{0}^{\infty} x \Re \left( e^{-i(x^2-x)} - e^{-i(x^2+x)} \right) e^{-sx^{2}} \, dx \\ &= \Re \int_{-\infty}^{\infty} x e^{-(s+i)x^2 + ix} \, dx. \end{align*}

By noting that

$$ z e^{-(s+i)z^2 + iz} = z \exp \left\{ -(s+i) \left( z - \tfrac{i}{2(s+i)} \right)^2 + \tfrac{i}{4(1-is)} \right\} $$

is an entire function with a nice vanishing speed as $ \left| \Re z \right| \to \infty$, we find that we can shift the contour of integration so that

\begin{align*} \int_{0}^{\infty} \sin x \sin \sqrt{x} \; e^{-sx} \, dx &= \Re \left[ e^{\frac{i}{4(1-is)}} \int_{-\infty}^{\infty} \left(x + \frac{i}{2(s+i)} \right) e^{-(s+i)x^2} \, dx \right] \\ &= \Re \left[ e^{\frac{i}{4(1-is)}} \frac{i}{2(s+i)} \int_{-\infty}^{\infty} e^{-(s+i)x^2} \, dx \right] \\ &= \Re \left[ e^{\frac{i}{4(1-is)}} \frac{i}{2(s+i)} \frac{\sqrt{\pi}}{\sqrt{s + i}} \right]. \end{align*}

Taking $s \to 0^{+}$, we find that in Abel summation sense,

\begin{align*} \int_{0}^{\infty} \sin x \sin \sqrt{x} \, dx &= \frac{\sqrt{\pi}}{2} \Re \left( e^{\frac{i-i\pi}{4}} \right) = \frac{\sqrt{\pi}}{2} \cos \left( \frac{1-\pi}{4} \right) = \frac{\sqrt{\pi}}{2} \sin \left( \frac{3\pi - 1}{4} \right) \end{align*}

as desired.


No, this integral is not convergent.

Consider $I(k):=\int_{0}^{\pi} \sin x\sin \sqrt{2k\pi +x}dx$. For $k$ large enough, $\sqrt{2k\pi +x}-\sqrt{2k\pi}, x\in [0,\pi]$ is small enough, so

$$\int_{0}^{\pi} \sin x\sin \sqrt{2k\pi +x}dx\sim \int_{0}^{\pi}\sin x \sin \sqrt{2k\pi }dx=2\sin \sqrt{2k\pi }\not\to 0.$$

  • $\begingroup$ I'm a bit confused, I was trying to follow what you have done. What does your integral say about the integral in the OP? I tried letting $x=y-2k\pi$ $\Rightarrow I(k)=\int_{2k\pi}^{(2k+1)\pi}{\sin(y)\sin\left(\sqrt{y}\right)}{dy}$, but I still don't see how this is related. $\endgroup$ – Graham Hesketh May 21 '13 at 12:37
  • $\begingroup$ @GrahamHesketh If the integral $F(u)=\int_{0}^{u}$ converges in the strict sense, then $I(k)=F((2k+1)\pi)-F(2k\pi)\to 0$. $\endgroup$ – Ma Ming May 21 '13 at 13:09
  • $\begingroup$ Sounds interesting, do you know the name for this or have a link abt it? $\endgroup$ – Graham Hesketh May 21 '13 at 13:19
  • $\begingroup$ @GrahamHesketh If $F(u)$ conveges as $u\to \infty$, then $F(u)$ is a en.wikipedia.org/wiki/Cauchy_sequence $\endgroup$ – Ma Ming May 21 '13 at 13:21

The integral cannot converge since the function $\sin{x}\sin{\sqrt{x}} $ doesn't converge through zero as x tends to infinity

EDIT : There is a mistake, this proof if wrong. Read the comments

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    $\begingroup$ This answer may be correct but the reasoning is flawed. For instance, $\int_0^\infty \sin{(x^2)}\,dx$ does converge though $\sin{(x^2)}$ doesn't converge to zero when $x \to \infty$. $\endgroup$ – Nick Strehlke May 21 '13 at 8:51

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