Re: Integrating tan(x)/x

In article <1177076761.537782.314520@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, *** says...

Hi all,

Does anybody know how the integral from 0 to infinity of tan(x)/x
get assigned the value of pi/2? I can't think of any way to naturally
assign a contour or of any way to manipulate Si(x) to yield it. Just
curious.

Thanks, ***

G. A. Edgar already gave the contour integral way of solving
the problem, but just for variety, here's the way that I did
it:

Note Integral from x=0 to infinity of tan(x)/x can
be written as

sum from n=0 to infinity of
Integral from x=0 to pi/2 tan(x + n pi)/(x+n pi) dx
+ Integral from x=pi/2 to pi of tan(x + n pi)/(x + n pi) dx

If we use tan(x+n pi) = tan(x) and
tan(pi - x) = - tan(x), we can rewrite this as

sum from n=0 to infinity of
Integral from x=0 to pi/2 tan(x)/(x+n pi) dx
+ Integral from x=0 to pi/2 of -tan(x)/((n+1) pi - x) dx

where in the second integral, I made the substitution
x --> pi - x. Assuming that it is permissable to interchange
summation with integration (which requires a proof, but I'm
not going to give it), this can be rewritten as

Integral from x=0 to pi/2 of tan(x) * f(x) dx

where f(x) = 1/x + 1/(pi + x) - 1/(pi - x) + 1/(2 pi + x) - 1/(2 pi - x) ...

= sum from n=-infinity to +infinity of 1/(x - n pi)

(where the infinite sum is the limit of the sum from n=-K to +K
as K --> infinity).

Since it is an alternating series, it will converge, provided that
x is not equal to a multiple of pi. What does it converge to?
Amazingly enough, it converges to cot(x). That's plausible, since
it has poles at all the right places (x = 0, +/- pi, +/- 2 pi, etc.).
To prove that f(x) = cot(x), one way is to start with the product
representation of sin(x), written in a special way:

sin(x) = x * 1/pi * (pi + x)
* 1/pi * (pi - x)
* 1/(2 pi) * (2 pi + x)
* 1/(2 pi) * (2 pi - x)
* ...

Then log(sin(x)) = log(x)
+ log(1/pi) + log(pi + x)
+ log(1/pi) + log(pi - x)
+ log(1/(2 pi)) + log(2pi + x)
+ log(1/(2 pi)) + log(2pi - x)
+ ...

Then d/dx log(sin(x))
= 1/x + 1/(pi + x) - 1/(pi - x) + 1/(2pi + x) - 1/(2pi - x) + ...

So cot(x) = 1/x + 1/(pi + x) - 1/(pi - x) + 1/(2pi + x) - 1/(2pi - x) + ...

--
Daryl McCullough
Ithaca, NY

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