Example (VI) – logarithms and the residue at infinity ContourLogs.png We seek to evaluate I = \int_0^3 \frac{x^{\frac{ 3}{4}} (3-x)^{\frac{1} {4}}} {5-x}\,dx. This requires a close study of f(z) = z^{\frac{3}{4}} (3-z)^{\frac{1} {4}}. We will construct f(z) so that it has a branch cut on [0, 3], shown in red in the diagram. To do this, we choose two branches of the logarithm, setting z^{\frac{3}{4}} = \exp \left (\frac{3}{4}\lo g(z) \right ) \quad \mbox{where} \quad -\pi \le \arg(z) < \pi and (3-z)^{\frac{1} {4}} = \exp \left (\frac{1}{4} \log(3-z) \right ) \quad \mbox{where} \quad 0 \le \arg(3-z) < 2\pi. The cut of z3/4 is therefore (−∞, 0] and the cut of (3−z)1/4 is (−∞, 3]. It is easy to see that the cut of the product of the two, i.e. f(z), is [0, 3], because f(z) is actually continuous across (−∞, 0). This is because when z = −r < 0 and we approach the cut from above, f(z) has the value r^{\frac{3}{4}} \exp(\tfrac{3 \pi i}{4}) (3+r)^{\frac{1} {4}} \exp(\tfrac{2 \pi i}{4}) = r^{\frac{3}{4}} (3+r)^{\frac {1} {4}} \exp(\tfrac{5 \pi i}{4}). When we approach from below, f(z) has the value r^{\frac{3}{4}} \exp(-\tfrac{3 \pi i}{4}) (3+r)^{\frac{1} {4}} \exp(\tfrac{0 \pi i}{4}) = r^{\frac{3}{4}} (3+r)^{\frac {1} {4}} \exp(-\tfrac{3 \pi i}{4}). But \exp(-\tfrac{3 \pi i}{4}) = \exp(\tfrac{5 \pi i}{4}), so that we have continuity across the cut. This is illustrated in the diagram, where the two black oriented circles are labelled with the corresponding value of the argument of the logarithm used in z3/4 and (3−z)1/4. We will use the contour shown in green in the diagram. To do this we must compute the value of f(z) along the line segments just above and just below the cut. Let z = r (in the limit, i.e. as the two green circles shrink to radius zero), where 0 ≤ r ≤ 3. Along the upper segment, we find that f(z) has the value r^{\frac{3}{4}} \exp(\tfrac{0 \pi i}{4}) (3-r)^{\frac{1} {4}} \exp(\tfrac{2 \pi i}{4}) = i \, r^{\frac{3}{4}} (3-r)^ {\frac{1} {4}} and along the lower segment, r^{\frac{3}{4}} \exp(3\tfrac{0 \pi i}{4}) (3-r)^{\frac{1} {4}} \exp(\tfrac{0 \pi i}{4}) = r^{\frac{3}{4}} (3-r)^{\frac {1} {4}}. It follows that the integral of \frac{f(z)}{5-z } along the upper segment is −iI in the limit, and along the lower segment, I. If we can show that the integrals along the two green circles vanish in the limit, then we also have the value of I, by the Cauchy residue theorem. Let the radius of the green circles be ρ, where ρ < 1/1000 and ρ → 0, and apply the ML-inequality. For the circle CL on the left, we find \left| \int_{C_L} \frac{f(z)}{5-z } dz \right| \le 2 \pi \rho \frac{\rho^{\fr ac{3}{4}} (3+\frac{1}{100 0})^{\frac{1}{4 }}}{5-\frac{1}{ 1000}} \in \mathcal{O} \left( \rho^{\frac {7}{ 4}} \right) \to 0. Similarly, for the circle CR on the right, we have \left| \int_{C_R} \frac{f(z)}{5-z } dz \right| \le 2 \pi \rho \frac{(3+\frac{ 1}{1000})^{\fra c{3}{4}} \rho^{\frac{1}{ 4}}}{2-\frac{1} {1000}} \in \mathcal{O} \left( \rho^{\frac {5}{ 4}} \right) \to 0. Now using the Cauchy residue theorem, we have (-i + 1) I = -2\pi i \left( \mathrm{Res}_{z =5} \frac{f(z)}{5- z } + \mathrm{Res}_{z =\infty} \frac{f(z)}{5-z } \right). where the minus sign is due to the clockwise direction around the residues. Using the branch of the logarithm from before, clearly \mathrm{Res}_{z =5} \frac{f(z)}{5-z } = - 5^{\frac{3} {4}} \exp \left (\tfrac{\log(-2 )}{4} \right). The pole is shown in blue in the diagram. The value simplifies to - 5^{\frac{3}{4}} \exp \left (\tfrac{\log(2) + \pi i}{4} \right ) = - \exp (\tfrac{\pi i}{4}) 5^{\frac{3}{4}} 2^{\frac {1}{4}} . We use the following formula for the residue at infinity: \mathrm{Res}_{z =\infty} h(z) = \mathrm{Res}_{z =0} \left[- \frac{1}{z^2} h\left(\frac{1} {z}\right)\righ t]. Substituting, we find \frac{1}{5-\fra c{1}{z}} = -z \left(1 + 5z + 5^2 z^2 + 5^3 z^3 + \cdots\right) and \left(\frac{1}{ z^3}\left (3-\frac{1}{z} \right )\right)^ {\frac {1}{4}} = \frac{1}{z} (3z-1)^{\frac{1 }{4}} = \frac {1}{z}\exp (\tfrac{\pi i}{4}) (1-3z)^{\frac{1 }{4}}, where we have used the fact that −1 = eiπ for the second branch of the logarithm. Next we apply the binomial expansion, obtaining \frac{1}{z} \exp(\tfrac{\pi i}{4}) \left( 1 - {\frac{1}{4} \choose 1} 3z + {\frac{1}{4} \choose 2} 3^2 z^2 - {\frac{1} {4} \choose 3} 3^3 z^3 + \cdots \right). The conclusion is that \mathrm{Res}_{z =\infty} \frac{f(z)}{5-z } = \exp (\tfrac {\pi i}{4}) \left (5 - \frac{3}{4} \right ) = \exp(\tfrac{\pi i} {4})\frac{17} {4}. Finally, it follows that the value of I is I = 2 \pi i \frac{\exp(\tfr ac{\pi i}{4})}{-1+i} \left(\frac{17} {4} - 5^{\frac{3}{4}} 2^{\frac{1}{4}} \right) = 2 \pi 2^{- \frac{1}{2} } \left(\frac{17} {4} - 5^{\frac{3}{4}} 2^ {\frac{1}{4}} \right) which yields I = \frac{\pi}{2\sq rt{2}} \left(17 - 5^{\frac{3}{4}} 2^ {\frac{9}{4}} \right) = \frac{\pi}{2\sq rt{2}} \left(17 - 40^ {\frac{3}{4} } \right). Integral representation Main article: Integral representation [icon] This section requires expansion. (November 2013) An integral representation of a function is an expression of the function involving a contour integral. Various integral representations are known for many special functions. Integral representations can be important for theoretical reasons, e.g. giving analytic continuation or functional equations, or sometimes for numerical evaluations. Hankels contour For example, the original definition of the Riemann zeta function \zeta(s) via a Dirichlet series, \sum_{n=1}^\inf ty\frac{1}{n^s} , is valid only for Re(s) > 1. But \zeta(s) = - \frac{\Gamma(1 - s)}{2 \pi i} \int\frac{(-t)^ {s-1}}{e^t - 1} dt , where the integration is done over the Hankel contour, is valid for all complex s.
Posted on: Mon, 05 Jan 2015 19:48:59 +0000
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