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  1. #1
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    An integral with a parameter

    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:
    The value of I_n does not depend on t.
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  2. #2
    Senior Member pankaj's Avatar
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    Plz give solution
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    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|.
    Last edited by halbard; October 10th 2009 at 07:54 AM. Reason: a,b,c
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    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:
    The value of I_n does not depend on 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 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}.
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  5. #5
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    Quote Originally Posted by DeMath View Post
    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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