\(\int_{0}^{+\infty}{\sin({x})\sin({\sqrt {x}\,})\,dx}\)

Calculus (Integrals, Series)
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Grigorios Kostakos
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\(\int_{0}^{+\infty}{\sin({x})\sin({\sqrt {x}\,})\,dx}\)

#1

Post by Grigorios Kostakos »

Prove that \(\displaystyle\int_{0}^{+\infty}{\sin({x})\sin\bigl({\sqrt {x}\,}\bigr)\,dx}=\frac{\sqrt{\pi}}{2}\sin\bigl({\tfrac{3\pi-1}{4}}\bigr)\,.\)

Proposed by Serafeim.
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Re: \(\int_{0}^{+\infty}{\sin({x})\sin({\sqrt {x}\,})\,dx}\)

#2

Post by r9m »

The integral is oscillating at infinity (does not converge). However, we can still define it in some limiting sense by taking $a \to 0$ here http://www.mathimatikoi.org/forum/viewt ... p=902#p902
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Re: \(\int_{0}^{+\infty}{\sin({x})\sin({\sqrt {x}\,})\,dx}\)

#3

Post by Grigorios Kostakos »

Very interesting!
r9m wrote:The integral is oscillating at infinity (does not converge).
r9m can you justify this?
r9m wrote:...However, we can still define it in some limiting sense by taking $a \to 0$ here http://www.mathimatikoi.org/forum/viewt ... p=902#p902
Can you explain in more details what do you mean?
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Re: \(\int_{0}^{+\infty}{\sin({x})\sin({\sqrt {x}\,})\,dx}\)

#4

Post by r9m »

To justify that the integral does not converge, we note that if $\displaystyle F(A) = \int_0^{A} f(x)\,dx$ converge as $A \to \infty$, then $F(A+\alpha) - F(A) \to 0$ as $A \to \infty$ (for any $\alpha >0$). But, this is not the case here, \begin{align*}\int_{2k\pi}^{(2k+1)\pi} \sin \left(\sqrt{x}\right)\sin x\,dx = \int_0^{\pi} \sin x \sin \left(\sqrt{x+2k\pi}\right)\,dx &\sim \int_0^{\pi} \sin x \sin \left(\sqrt{2k\pi}\right)\,dx \\&= 2\sin \left(\sqrt{2k\pi}\right) \not\to 0\end{align*} as $k \to \infty$.

However, like we saw in the previous post the integral $\displaystyle \int_0^{\infty} e^{-ax}\sin \left(\sqrt{x}\right)\sin x\,dx$ converges when $a > 0$ but we will see now that convergence fails at $a = 0$.

In general $\displaystyle \lim\limits_{a \to 0^{+}} \int_0^{\infty} e^{-ax} f(x)\,dx \neq \int_0^{\infty} f(x)\,dx$ (the passage of limit inside the integral is not justified). If $f$ is absolutely integrable, then passage of the limit can be justified by Lebesgue Dominated convergence theorem (each $f_a(x) = e^{-ax}f(x)$ for $a > 0$ is bounded above by $|f_a(x)| < |f(x)|$ with $\displaystyle \lim\limits_{a\to 0^{+}} f_a(x) = f(x)$ pointwise and since $f \in L^{1}(0,\infty)$, we have $\displaystyle \lim\limits_{a \to 0^{+}} \int_0^{\infty} f_a(x) = \int_0^{\infty} f(x)$). However, even if $f$ is not absolutely integrable we can ensure passage of limit v.i.a. uniform convergence of integrals, which requires for all $\epsilon > 0$ to have a $N(\epsilon) > 0$ such that $\displaystyle \left| \int_{N(\epsilon)}^{\infty} e^{-ax}f(x)\,dx \right| < \epsilon$ for each $a \in [0,\delta]$.

To see how $\displaystyle F(A) = \int_0^{A} e^{-ax}f(x)\,dx$ behaves in our case $\displaystyle f(x) = \sin \left(\sqrt{x}\right)\sin x$, we integrate by parts once, \begin{align*} F(A) &= -e^{-aA}\sin \left(\sqrt{A}\right)\cos A + \frac{1}{2}\int_0^{A}\frac{e^{-ax} \cos x \cos \left(\sqrt{x}\right)}{\sqrt{x}} - ae^{-ax} \sin\left(\sqrt{x}\right)\cos x \,dx \\&= -e^{-aA}\sin \left(\sqrt{A}\right)\cos A + \int_0^{\infty} e^{-ax^2}\cos \left(x^2\right)\cos x\,dx + o(1) + O(a) \\&= -e^{-aA}\sin \left(\sqrt{A}\right)\cos A + \frac{1}{2}\Re \left(e^{-\frac{1}{4(a+i)}}\sqrt{\frac{\pi}{a+i}}\right) + o(1) +O(a)\\&= -e^{-aA}\sin \left(\sqrt{A}\right)\cos A + \frac{\sqrt{\pi}}{2(a^2+1)^{1/4}} \cos \left(\frac{1}{4(a^2+1)} - \frac{1}{2}\arctan \frac{1}{a}\right) + o(1) +O(a)\end{align*}

Thus, when $a \to 0^{+}$, $\displaystyle F(A) = -\sin \left(\sqrt{A}\right)\cos A + \frac{\sqrt{\pi}}{2} \sin \left(\frac{3\pi - 1}{4}\right) + o(1)$, which clearly oscillates as $A \to \infty$. But it is well defined when $a > 0$, in which case the oscillating part vanishes as $A \to \infty$, which is why I mentioned that we can talk about convergence in a limiting sense.
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Re: \(\int_{0}^{+\infty}{\sin({x})\sin({\sqrt {x}\,})\,dx}\)

#5

Post by mathofusva »

Some Remarks:

By the substitution $x = t^2$, the proposed integral becomes
$$2\,\int_0^\infty\,t\sin(t^2)\sin t\,dt.$$
In 1851, Cauchy obtained
$$ \int_0^\infty\,\sin(t^2)\cos(pt)\,dt = \frac{1}{2}\,\sqrt{\frac{\pi}{2}}\,\left(\cos\frac{p^2}{4} - \sin\frac{p^2}{4}\right).$$
He then differentiated under the integral sign with respect to $p$ yielding
$$ \int_0^\infty\,t\sin(t^2)\sin(pt)\,dt = \frac{p}{4}\,\sqrt{\frac{\pi}{2}}\,\left(\cos\frac{p^2}{4} + \sin\frac{p^2}{4}\right). \tag{1}$$
This formula has subsequently been reproduced and still appears in standard tables today. However, in the paper ``Some divergent trigonometric integrals", Amer. Math. Monthly, 108(2001) 432-436, Talvila proved that the integral in $(1)$ is divergent!
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Re: \(\int_{0}^{+\infty}{\sin({x})\sin({\sqrt {x}\,})\,dx}\)

#6

Post by Tolaso J Kos »

mathofusva wrote: This formula has subsequently been reproduced and still appears in standard tables today. However, in the paper ``Some divergent trigonometric integrals", Amer. Math. Monthly, 108(2001) 432-436, Talvila proved that the integral in $(1)$ is divergent!
Very interesting. Is the issue you are referring to of free access? If so, may you point us with a link?

T.
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Re: \(\int_{0}^{+\infty}{\sin({x})\sin({\sqrt {x}\,})\,dx}\)

#7

Post by mathofusva »

For example, you can find the following formula

$$\int_0^\infty\,x\sin(ax^2)\sin(2bx)\,dx = \frac{b}{2a}\sqrt\frac{\pi}{2a}\left(\cos\frac{b^2}{a} + \sin\frac{b^2}{a}\right)$$

in Gradshteyn & Ryzhik's Table of Integrals, Series, and Products as Formula 3.851.1, on page 500 (5th edition).
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