An integral with a parameter

**NonCommAlg** · October 4th 2009, 03:02 AM

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

Spoiler:

**pankaj** · October 10th 2009, 06:27 AM

Plz give solution

**halbard** · October 10th 2009, 07:52 AM

Originally Posted by NonCommAlg

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

More generally, evaluate $\int_0^\infty\frac{\sin(ax)}{x^{n+1}}\prod_{i=1}^n \sin(b_ix)\prod_{j=1}^m\cos(c_jx)\,\mathrm dx$ where $a>\sum_{i=1}^n|b_i|+\sum_{j=1}^m|c_j|$ .

**DeMath** · December 12th 2009, 11:07 AM

Originally Posted by NonCommAlg

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

Spoiler:

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

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

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

${I_1} = \int\limits_0^{a} {\frac{{\sin (tx)}}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} dx} = \left\{ \begin{gathered}x = \frac{y}{t}, \hfill \\dx = \frac{{dy}}{t} \hfill \\ \end{gathered} \right\} = {t^n}\int\limits_0^{at} {\frac{{\sin y}}{{{y^{n + 1}}}}\prod\limits_{k = 1}^n {\sin \left( {\frac{{ky}}{t}} \right)} dy} =$ ${\color{white}a^3t \to 0,ift \to \infty}(1)$

$= {t^n}\int\limits_0^{at} {\frac{{\sin y}}{y}\prod\limits_{k = 1}^n {\left( {\frac{{ky}}{t} + O\left( {{{\left( {\frac{y}{t}} \right)}^3}} \right)} \right)\frac{{dy}}{{{y^n}}}} } = n!\int\limits_0^{at} {\frac{{\sin y}}{y}\,dy} + O\left( {{a^3}t} \right).$

Now we impose a condition on the parameter: a $^3t \to 0\,\,\text{ if }\,t \to \infty$ .
Now we estimate the integral on the interval $[a,\,\infty)$ :

$I_2= \int\limits_a^\infty {\frac{{\sin (tx)}}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)}\,dx} =$ $\left. { - \frac{{\cos \left( {tx} \right)}}{{t{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} } \right|_{x = a}^{x \to \infty } + \frac{1}{t}\int\limits_a^\infty {\frac{d}{{dx}}\left[ {\frac{1}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} } \right]\cos(tx)\,dx} .$

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

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

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

**NonCommAlg** · December 12th 2009, 02:41 PM

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

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

${I_1} = \int\limits_0^{a} {\frac{{\sin (tx)}}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} dx} = \left\{ \begin{gathered}x = \frac{y}{t}, \hfill \\dx = \frac{{dy}}{t} \hfill \\ \end{gathered} \right\} = {t^n}\int\limits_0^{at} {\frac{{\sin y}}{{{y^{n + 1}}}}\prod\limits_{k = 1}^n {\sin \left( {\frac{{ky}}{t}} \right)} dy} =$ ${\color{white}a^3t \to 0,ift \to \infty}(1)$

$= {t^n}\int\limits_0^{at} {\frac{{\sin y}}{y}\prod\limits_{k = 1}^n {\left( {\frac{{ky}}{t} + O\left( {{{\left( {\frac{y}{t}} \right)}^3}} \right)} \right)\frac{{dy}}{{{y^n}}}} } = n!\int\limits_0^{at} {\frac{{\sin y}}{y}\,dy} + O\left( {{a^3}t} \right).$

Now we impose a condition on the parameter: a $^3t \to 0\,\,\text{ if }\,t \to \infty$ .
Now we estimate the integral on the interval $[a,\,\infty)$ :

$I_2= \int\limits_a^\infty {\frac{{\sin (tx)}}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)}\,dx} =$ $\left. { - \frac{{\cos \left( {tx} \right)}}{{t{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} } \right|_{x = a}^{x \to \infty } + \frac{1}{t}\int\limits_a^\infty {\frac{d}{{dx}}\left[ {\frac{1}{{{x^{n + 1}}}}\prod\limits_{k = 1}^n {\sin (kx)} } \right]\cos(tx)\,dx} .$

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

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

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

well, it's easier than that!

we first show, by induction over $n,$ that the value of the integral, $I_n,$ does not depend on $t.$ for n = 1, we have t > 1. then:

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

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

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

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

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

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

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

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

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