Monday Math 29

Consider the function , defined by the integral: . Can we find a formula for in terms of known functions?

First, we note that for , we have the series . Thus . Making the change of variable , we find that for , . Now, setting , we see , for .

Thus
.

Now, for the integral in each term, , we make the substitution . Then and , so our integral becomes:
,
and you should recognize that last integral as the gamma function of x.
Therefore, we find:
,
and that last sum is also familiar: it is the Riemann zeta function.
Thus, we see that

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5 Responses to “Monday Math 29”

  1. Physics Friday 33: Blackbody Radiation (Part 2) « Twisted One 151’s Weblog Says:

    […] . Using the u substitution , we find , , we find . Now, to find the remaining integral, I showed in Monday Math 29 that , we use x=4 to see that . Now, , and we found in Monday Math 26 that , so , and thus , where […]

  2. Monday Math 35 « Twisted One 151’s Weblog Says:

    […] Math 35 In a previous post, we found that . What, then, is ? Previously, we found that for , . Similarly, as , we see that […]

  3. Monday Math 55 « Twisted One 151’s Weblog Says:

    […] can be seen as between and x=1, u>0. So, reversing the order of the double integral, We found previously that , and the gamma function is defined as ; this means This is called Hadjicostas’s […]

  4. Monday Math 77 « Twisted One 151’s Weblog Says:

    […] make the u-substitution u=2x to get and renaming the variable back to x, we get . Now, we showed here that . Thus, our integral is […]

  5. Monday Math 81 « Twisted One 151’s Weblog Says:

    […] see . Now, , so the above becomes . Now, along with our formula that for positive integer n (see here), we can derive a couple of interesting results. First, consider . Expanding the sine in its […]

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