Re: complex integral

In article <1138423818.444739.87060@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"tanabai" <diakam@xxxxxx> wrote:
>Thanks a lot. The minus is really false. I will try it to calculate
>without computer. But it is good to know that there is a possible
>result.

I have put an image of your original TeX post at

<http://www.whim.org/nebula/math/images/others/tanabai.gif>

There is nothing keeping us from passing to the limits and setting
both \delta and \epsilon to 0. Therefore, as originally posted, the
integral is 0. However, if as assumed by Paul Abbott, you had
intended a '+' in place of the '-', then both summands are equal and
the integral becomes

1 |\T dn
-- | ------------------------
pi \| 0 sqrt(n) (1 + ik sqrt(n))

2 |\sqrt(T) dx 2
= -- | --------- [ n = x ]
pi \| 0 (1 + ikx)

2
= ----- log(1 + ik sqrt(T))
ik pi

2 2
= ----- (log(sqrt(1 + k T)) + i atan(k sqrt(T)))
ik pi

1 2
= ---- (2 atan(k sqrt(T)) - i log(1 + k T))
k pi

The result you were looking for is just the real part of

1 |\T dn
-- | ------------------------
pi \| 0 sqrt(n) (1 + ik sqrt(n))

1 |\T (1 - ik sqrt(n)) dn
= -- | -------------------
pi \| 0 sqrt(n) (1 + k^2 n)

Rob Johnson <rob@xxxxxxxxxxxxxx>
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