Analysis Chapter7
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7 App lica tion to rea l integra ls and ser ies Lecture 18 We first consider a particularly simple example with few technical difficulties so that the line of argument is clear. Example 7.1. Calculate I = ∞ −∞ 1 1 + x 2 dx. Put x = tan θ. Then I = π 2 − π 2 1 sec 2 θ sec 2 θ dθ = π. Next we to do this exercise using the calculus of residues. Let R > 0. Con- sider the contour which we will denote throughout the remaining lectures by Γ R :this consists of running along the real axis from −R to R, then around the semicircle centered at 0 of radius R in the upper half plane to return to −R. (See diagram). This contour will be used in many examples. Then Γ R dz 1 + z 2 = 2πi (re sidueof 1 1 + z 2 at z = i) = 2πi 1 2i = π. To estimate the contribution around the semicircle, put z = Re it , t ∈ [0, π]. Then π 0 Rie it dt 1 + ( Re it ) 2 ≤ π 0 R R 2 − 1 dt = π R R 2 − 1 = O( 1 R ) which → 0 as R →∞. Then Γ R 1 1 + z 2 dz → ∞ −∞ dx 1 + x 2 and we have shown that ∞ −∞ dx 1 + x 2 = π. Example 7.2. Calculate ∞ −∞ 1 + x 2 1 + x 4 dx. To find the roots of z 4 +1, we note that z 8 − 1 = (z 4 − 1)(z 4 + 1), so the roots of z 4 + 1 are four of the eighth roots of 1, namely, e i π 4 , e i π 4 3 , e i π 4 5 , e i π 4 7 , points at which z 4 + 1 has simple zeros, so the integrand has simple poles. 57
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