Singularities/Residues/ Applications 7131
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9.
0(tan or dO =^1 krrsec^1 / 2 mr, -1 < a< 1 (Hint: Let x = btanO
in problem 2.)1
(^1) x (^1) -a(l - x)a 7ra(l - a)
- ( ) 3 dx = 23. , -1 <a< 2 [Let (1 - x)/x = t.]
0 1 + x -a sma7r
Type VIII. Jo"° f(x)(lnx)n dx, where n is a positive integer and f(x)
satisfies the following conditions:- f(x) has an extension J(z) that is meromorphic in C.
- f(z) has no real poles.
3. f(-z) = f(z).
- f(z) = 0(1/z°'), a > 1, as z ---t oo.
Consider Ia+ f(z)(logz)n dz, where c+ is the contour shown in
Fig. 9.22, namely, c+ = [-R, -r] + 'Y-+ [r, R] + r+, 0 < r < R,
and for z = JzJei^9 (z =f. 0), log z is defined by log z = ln Jzl + iO, with
-1/i7r < 0::; 3fi7r. The reader will observe that by using this branch of the
logarithm we have log x = ln x for x > 0, and log x = ln Jxl + i7r for x < 0.
Of course, the indentation about 0 is made in order to bypass the branch
point oflogz at z = 0. As we have noted before, condition (4) implies that
f admits a finite number of poles in C, so by choosing r small enough and
R sufficiently large, all poles of f (if any) on Im z > 0 will be enclosed by
c+. Assuming that there are some poles bk ( k = 1, ... , m) we have, by
the residue theorem,
1: f(x)(lnlxl +i7rrdx + j f(z)(logztdz
'Y-b~ f+/,
R xFig. 9.22