An integral with a parameter

• Oct 4th 2009, 03:02 AM
NonCommAlg
An integral with a parameter
Let $\displaystyle n \in \mathbb{N}$ and $\displaystyle \frac{n(n+1)}{2} < t \in \mathbb{R}.$ Evaluate $\displaystyle I_n = \int_0^{\infty} \frac{\sin (x) \sin (2x) \ \cdots \ \sin(nx) \sin (tx)}{x^{n+1}} \ dx.$

Spoiler:
The value of $\displaystyle I_n$ does not depend on $\displaystyle t.$
• Oct 10th 2009, 06:27 AM
pankaj
Plz give solution
• Oct 10th 2009, 07:52 AM
halbard
Quote:

Originally Posted by NonCommAlg
Let $\displaystyle n \in \mathbb{N}$ and $\displaystyle \frac{n(n+1)}{2} < t \in \mathbb{R}.$ Evaluate $\displaystyle I_n = \int_0^{\infty} \frac{\sin (x) \sin (2x) \ \cdots \ \sin(nx) \sin (tx)}{x^{n+1}} \ dx.$

More generally, evaluate $\displaystyle \int_0^\infty\frac{\sin(ax)}{x^{n+1}}\prod_{i=1}^n \sin(b_ix)\prod_{j=1}^m\cos(c_jx)\,\mathrm dx$ where $\displaystyle a>\sum_{i=1}^n|b_i|+\sum_{j=1}^m|c_j|$.
• Dec 12th 2009, 11:07 AM
DeMath
Quote:

Originally Posted by NonCommAlg
Let $\displaystyle n \in \mathbb{N}$ and $\displaystyle \frac{n(n+1)}{2} < t \in \mathbb{R}.$ Evaluate $\displaystyle I_n = \int_0^{\infty} \frac{\sin (x) \sin (2x) \ \cdots \ \sin(nx) \sin (tx)}{x^{n+1}} \ dx.$

Spoiler:
The value of $\displaystyle I_n$ does not depend on $\displaystyle t.$

Of course, my solution is not the most elegant, but I think it is quite reasonable.

We define the independence of the integral from $\displaystyle t$, for which we use the fact that the function of the form $\displaystyle \frac{\sin(ax)}{x}$ up to $\displaystyle a$ constant is the Fourier transform of unit step height in the interval $\displaystyle [-a,\,a]$. Consider this integral as the value at $\displaystyle 0$ inverse Fourier transform of the product of the functions $\displaystyle \frac{\sin(tx)}{x^{n + 1}}\,\prod\limits_{k = 1}^n{\sin(kx)}$, but the contraction of the first n steps will be of compact support on the interval $\displaystyle \left[-\frac{n(n +1)}{2},\,\frac{n(n +1)}{2}\right]$. Therefore, if $\displaystyle t>\frac{n(n +1)}{2}$, when calculating the convolution of this function with the last step, the width of $\displaystyle 2t$ at point $\displaystyle 0$, we simply calculate the area under the graph of the convolution of the first $\displaystyle n$ steps, but this area does not depend on $\displaystyle t>\frac{n(n +1)}{2}$.

We divide this integral $\displaystyle I(t)=I_1+I_2$ into two intervals $\displaystyle [0,\,a] \wedge [a,\,\infty)$. Parameter $\displaystyle a=a(t)$ we will suggest later, based on the conditions: $\displaystyle a(t)\to 0\,\text{ if }\,t\to\infty$. Now perform the substitution $\displaystyle x=\frac{y}{t}$ and estimate this integral on the interval $\displaystyle [0,\,a]$:

$\displaystyle {I_1} = \int\limits_0^{a} {\frac{{\sin (tx)}}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} dx} = \left\{ \begin{gathered}x = \frac{y}{t}, \hfill \\dx = \frac{{dy}}{t} \hfill \\ \end{gathered} \right\} = {t^n}\int\limits_0^{at} {\frac{{\sin y}}{{{y^{n + 1}}}}\prod\limits_{k = 1}^n {\sin \left( {\frac{{ky}}{t}} \right)} dy} =$$\displaystyle {\color{white}a^3t \to 0,ift \to \infty}(1) \displaystyle = {t^n}\int\limits_0^{at} {\frac{{\sin y}}{y}\prod\limits_{k = 1}^n {\left( {\frac{{ky}}{t} + O\left( {{{\left( {\frac{y}{t}} \right)}^3}} \right)} \right)\frac{{dy}}{{{y^n}}}} } = n!\int\limits_0^{at} {\frac{{\sin y}}{y}\,dy} + O\left( {{a^3}t} \right). Now we impose a condition on the parameter: a\displaystyle ^3t \to 0\,\,\text{ if }\,t \to \infty. Now we estimate the integral on the interval \displaystyle [a,\,\infty): \displaystyle I_2= \int\limits_a^\infty {\frac{{\sin (tx)}}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)}\,dx} =$$\displaystyle \left. { - \frac{{\cos \left( {tx} \right)}}{{t{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} } \right|_{x = a}^{x \to \infty } + \frac{1}{t}\int\limits_a^\infty {\frac{d}{{dx}}\left[ {\frac{1}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} } \right]\cos(tx)\,dx} .$

From this representation $\displaystyle I_2$ and given the inequality $\displaystyle \left| \frac{\sin x}{x}\right| \leqslant 1$, we obtain an estimate $\displaystyle \left|I_n \right|\leqslant\frac{C}{at}$. And now we impose the condition $\displaystyle at \to \infty\,\text{ if }\,t\to\infty$. Let $\displaystyle a = a(t) = t^{-\alpha},\,\alpha > 0$. Then the condition $\displaystyle a^3t \to 0$ that means $\displaystyle 1 - 3\alpha < 0, \, \alpha >\frac{1}{3}$, but the condition $\displaystyle at\to\infty$ means $\displaystyle 1-\alpha > 0, \,\alpha < 1$. Thus, when $\displaystyle \frac{1}{3}<\alpha<1$, all conditions are met.

So, on the basis of $\displaystyle (1)$, it follows that

$\displaystyle \mathop{\lim }\limits_{t \to \infty}I(t) = \frac{\pi n!}{2}.$
• Dec 12th 2009, 02:41 PM
NonCommAlg
Quote:

Originally Posted by DeMath
Of course, my solution is not the most elegant, but I think it is quite reasonable.

We define the independence of the integral from $\displaystyle t$, for which we use the fact that the function of the form $\displaystyle \frac{\sin(ax)}{x}$ up to $\displaystyle a$ constant is the Fourier transform of unit step height in the interval $\displaystyle [-a,\,a]$. Consider this integral as the value at $\displaystyle 0$ inverse Fourier transform of the product of the functions $\displaystyle \frac{\sin(tx)}{x^{n + 1}}\,\prod\limits_{k = 1}^n{\sin(kx)}$, but the contraction of the first n steps will be of compact support on the interval $\displaystyle \left[-\frac{n(n +1)}{2},\,\frac{n(n +1)}{2}\right]$. Therefore, if $\displaystyle t>\frac{n(n +1)}{2}$, when calculating the convolution of this function with the last step, the width of $\displaystyle 2t$ at point $\displaystyle 0$, we simply calculate the area under the graph of the convolution of the first $\displaystyle n$ steps, but this area does not depend on $\displaystyle t>\frac{n(n +1)}{2}$.

We divide this integral $\displaystyle I(t)=I_1+I_2$ into two intervals $\displaystyle [0,\,a] \wedge [a,\,\infty)$. Parameter $\displaystyle a=a(t)$ we will suggest later, based on the conditions: $\displaystyle a(t)\to 0\,\text{ if }\,t\to\infty$. Now perform the substitution $\displaystyle x=\frac{y}{t}$ and estimate this integral on the interval $\displaystyle [0,\,a]$:

$\displaystyle {I_1} = \int\limits_0^{a} {\frac{{\sin (tx)}}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} dx} = \left\{ \begin{gathered}x = \frac{y}{t}, \hfill \\dx = \frac{{dy}}{t} \hfill \\ \end{gathered} \right\} = {t^n}\int\limits_0^{at} {\frac{{\sin y}}{{{y^{n + 1}}}}\prod\limits_{k = 1}^n {\sin \left( {\frac{{ky}}{t}} \right)} dy} =$$\displaystyle {\color{white}a^3t \to 0,ift \to \infty}(1) \displaystyle = {t^n}\int\limits_0^{at} {\frac{{\sin y}}{y}\prod\limits_{k = 1}^n {\left( {\frac{{ky}}{t} + O\left( {{{\left( {\frac{y}{t}} \right)}^3}} \right)} \right)\frac{{dy}}{{{y^n}}}} } = n!\int\limits_0^{at} {\frac{{\sin y}}{y}\,dy} + O\left( {{a^3}t} \right). Now we impose a condition on the parameter: a\displaystyle ^3t \to 0\,\,\text{ if }\,t \to \infty. Now we estimate the integral on the interval \displaystyle [a,\,\infty): \displaystyle I_2= \int\limits_a^\infty {\frac{{\sin (tx)}}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)}\,dx} =$$\displaystyle \left. { - \frac{{\cos \left( {tx} \right)}}{{t{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} } \right|_{x = a}^{x \to \infty } + \frac{1}{t}\int\limits_a^\infty {\frac{d}{{dx}}\left[ {\frac{1}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} } \right]\cos(tx)\,dx} .$

From this representation $\displaystyle I_2$ and given the inequality $\displaystyle \left| \frac{\sin x}{x}\right| \leqslant 1$, we obtain an estimate $\displaystyle \left|I_n \right|\leqslant\frac{C}{at}$. And now we impose the condition $\displaystyle at \to \infty\,\text{ if }\,t\to\infty$. Let $\displaystyle a = a(t) = t^{-\alpha},\,\alpha > 0$. Then the condition $\displaystyle a^3t \to 0$ that means $\displaystyle 1 - 3\alpha < 0, \, \alpha >\frac{1}{3}$, but the condition $\displaystyle at\to\infty$ means $\displaystyle 1-\alpha > 0, \,\alpha < 1$. Thus, when $\displaystyle \frac{1}{3}<\alpha<1$, all conditions are met.

So, on the basis of $\displaystyle (1)$, it follows that

$\displaystyle \mathop{\lim }\limits_{t \to \infty}I(t) = \frac{\pi n!}{2}.$

well, it's easier than that! (Nod) we first show, by induction over $\displaystyle n,$ that the value of the integral, $\displaystyle I_n,$ does not depend on $\displaystyle t.$ for n = 1, we have t > 1. then:

$\displaystyle I_1=\int_0^{\infty} \frac{\sin(x) \sin (tx)}{x^2} \ dx=\frac{1}{2} \int_0^{\infty} \frac{\cos (t-1)x - \cos(t+1)x}{x^2} \ dx$

$\displaystyle =\frac{1}{2} \int_0^{\infty} \frac{1}{x} \int_{t-1}^{t+1} \sin(yx) \ dy \ dx = \frac{1}{2} \int_{t-1}^{t+1} \int_0^{\infty} \frac{\sin (yx)}{x} \ dx \ dy=\frac{\pi}{2}.$

now let $\displaystyle n > 1$ and suppose that the claim is true for $\displaystyle I_{n-1}.$ then, exactly the same as above, we'll have:

$\displaystyle I_n=\frac{1}{2} \int_0^{\infty} \frac{\sin(x) \sin(2x) \cdots \sin[(n-1)x]}{x^n} \int_{t-n}^{t+n} \sin(yx) \ dy \ dx$

$\displaystyle =\frac{1}{2} \int_{t-n}^{t+n} \int_0^{\infty} \frac{\sin(x) \sin(2x) \cdots \sin[(n-1)x] \sin(yx)}{x^n} \ dx \ dy$ *

$\displaystyle =\frac{1}{2}I_{n-1} \int_{t-n}^{t+n} dy=nI_{n-1}.$

so we also proved that $\displaystyle I_n = nI_{n-1}, \ n \geq 2,$ which, with $\displaystyle I_1=\frac{\pi}{2},$ gives us $\displaystyle I_n = \frac{\pi n!}{2}. \ \Box$

* note that $\displaystyle y \geq t-n > \frac{n(n+1)}{2} - n = \frac{n(n-1)}{2}.$