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By Leif Mejlbro

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16 Given the function f (z) = z2 . z4 + z2 + 1 1) Find all isolated singularities of f in C, and specify their types. 2) Prove by using the calculus of residues that the improper integral +∞ x4 0 x2 dx + x2 + 1 π is convergent of the value √ . 2 3 One may use that f (x) is an even function. 1) First note that z2 − 1 z4 + z2 + 1 = z6 − 1 = 0 for z = exp i pπ , 3 p ∈ Z. When we again remove the roots z = ±1 of the auxiliary factor z 2 − 1, we see that the simple poles are √ √ π 1 3 2π 1 3 exp i = +i , exp i =− +i , 3 2 2 3 2 2 √ 1 3 = −i , 2 2 π exp −i 3 2π exp −i 3 √ 1 3 .

3 The denominator can be factorized in the following way 1 + x6 = 1 + x2 = x2 + 1 x4 − x2 + 1 = 1 + x2 x4 + 2x3 + 1 − 3x2 √ √ 2 x2 + 1 − ( 3 x)2 = x2 + 1 x2 + 3 x + 1 x2 − √ 3x + 1 , so we can in principle decompose the integrand and then integrate in the usual way known from Calculus. However, the coefficients clearly show that this will be very difficult to carry through. ♦ (b) Here we must not forget what we learned in the “kindergarten”: +∞ −∞ x2 1 dx = 1 + x6 3 +∞ −∞ d x3 2 1 + (x3 ) = 1 Arctan x3 3 +∞ −∞ = π .

Z→i dz 1 (z + i)3 (−3)(−4) 12πi 12πi 3π = = = . 4 Prove that +∞ (a) −∞ √ π 2 x2 dx = , x4 + 1 2 +∞ (b) −∞ 4π 2π x−1 dx = sin . 5 5 x5 − 1 x2 has a zero of second order at ∞, and no singularity on the x-axis. The +1 poles in the upper half plane, (a) The integrand 1+i z1 = √ 2 x4 and z2 = −1 + i √ , 2 are both simple with z1 · z2 = −1, so +∞ −∞ x2 dx = 2πi res x4 + 1 = 2πi = − π 2 lim z→z1 z2 ; z1 z4 + 1 + res z2 z2 + lim 3 z→z2 4z 3 4z 1 + i −1 + i √ + √ 2 2 z2 ; z2 z4 + 1 2πi 4 1 1 + z2 z1 πi =− (z1 + z2 ) 2 √ πi 2i π π 2 =− ·√ = √ = .

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