Let and Evaluate

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- Oct 4th 2009, 04:02 AMNonCommAlgAn integral with a parameter
Let and Evaluate

__Spoiler__: - Oct 10th 2009, 07:27 AMpankaj
Plz give solution

- Oct 10th 2009, 08:52 AMhalbard
- Dec 12th 2009, 12:07 PMDeMath
Of course, my solution is not the most elegant, but I think it is quite reasonable.

We define the independence of the integral from , for which we use the fact that the function of the form up to constant is the Fourier transform of unit step height in the interval . Consider this integral as the value at inverse Fourier transform of the product of the functions , but the contraction of the first n steps will be of compact support on the interval . Therefore, if , when calculating the convolution of this function with the last step, the width of at point , we simply calculate the area under the graph of the convolution of the first steps, but this area does not depend on .

We divide this integral into two intervals . Parameter we will suggest later, based on the conditions: . Now perform the substitution and estimate this integral on the interval :

Now we impose a condition on the parameter: a .

Now we estimate the integral on the interval :

From this representation and given the inequality , we obtain an estimate . And now we impose the condition . Let . Then the condition that means , but the condition means . Thus, when , all conditions are met.

So, on the basis of , it follows that

- Dec 12th 2009, 03:41 PMNonCommAlg
well, it's easier than that! (Nod) we first show, by induction over that the value of the integral, does not depend on for n = 1, we have t > 1. then:

now let and suppose that the claim is true for then, exactly the same as above, we'll have:

*****

so we also proved that which, with gives us

*****note that