36.6. EVALUATION OF IMPROPER INTEGRALS 745

It is left as an exercise to verify that cotπz is bounded on this contour and that therefore,IN → 0 as N → ∞. Now compute the residues of the integrand at ±α and at n where|n|< N+ 1

2 for n an integer. These are the only singularities of the integrand in this contourand therefore, IN can be obtained by using these. First consider the residue at ±α . Theseare obviously poles of order 1 and so to get the one at α, you take

limz→α

(z−α)π cosπz(α2− z2)sinπz

= limz→α

−π cosπz(α + z)sinπz

=−π cosπα

2α sinπα

You get the same thing at −α . Next consider the residue at n. If you consider the powerseries, you will see that this should also be a pole of order 1. Thus it is

limz→n

(z−n)π cosπz(α2− z2)sinπz

= limz→n

π cosπz− (z−n)π2 sin(πz)−2zsinπz+(α2− z2)π cos(πz)

=π (−1)n

(α2−n2)π (−1)n =1

α2−n2

Therefore,

0 = limN→∞

IN = limN→∞

2πi

[N

∑n=−N

1α2−n2 −

π cotπα

α

]which establishes the following formula of Mittag Leffler.

limN→∞

N

∑n=−N

1α2−n2 =

π cotπα

α.

Writing this in a slightly nicer form, we obtain 36.10.The next example illustrates the technique of a branch cut.

Example 36.6.8 For p ∈ (0,1) , find∫

∞

0xp−1

1+x dx. This example illustrates the use of some-thing called a branch cut. The idea is you need to pick a single determination of zp−1

which converges to xp−1 for x real and z getting close to x. It will make use of the followingcontour. In this contour, the radius of the large circle is R and the radius of the small oneis r. The angle between the straight lines and the x axis is ε . Denote this contour by γR,r,ε .

Choose a branch of the logarithm of the form log(z) = ln |z|+ iA(z) where A(z) is theangle of z in (0,2π). Thus

zp−1 = e(p−1)(ln|z|+iA(z))

The straight lines, the one on top. reiε + t(Reiε

)= z, t ∈ [0,1].