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Z (z 2 + 1) 1) Prove that π f (z) dz = √ (1 − i). 2 Γr,R 2) Prove that the integrals of f along the half circles IIR and IVr tend to 0 when R tends to ∞, and r tends to 0. 3) Prove that the integral +∞ 0 1 √ dx x (x2 + 1) is convergent and find its value. com 40 Complex Funktions Examples c-7 Improper integrals in general 1) The only singularity of f (z) inside Γr,R is the simple pole z = i, so it follows by Cauchy’s residuum theorem that f (z) dz Γr,R = 2πi res 1 √ ,i z (z 2 + 1) = 2πi lim √ z→i 1 1 = 2πi · 1+i z · 2z √ · 2i 2 π 1−i = π · √ = √ (1 − i).

1 4 t2 Finally, by separating the real and the imaginary parts, +∞ 0 x2 ln x dx = 2 4 x −1 +∞ og 0 ln x dx = 0. 7 Given the function f (z) = ez . 1 + e4z (1) Find all the isolated singularities of f in C. Determine the type of each of them and their residuum. Given for each r1 > 0 and r2 > 0 the closed curve γr1 ,r2 = Ir1 ,r2 + IIr2 + IIIr1 ,r2 + IVr1 (cf. the figure), which form the boundary of the domain Ar1 ,r2 = {z ∈ C | −r1 < Re(z) < r2 and 0 < Im(z) < π}. 6 4 2 –10 –5 5 10 –2 Figure 5: The curve γr1 ,r2 with the direction given on Ir1 ,r2 = [−r1 , r2 ] and IIIr1 ,r2 .

Com 56 Complex Funktions Examples c-7 Improper integral, where the integrand is a rational function so by insertion, +∞ x2 dx 2 (x2 + 1) (x2 + 4) −∞ = 2πi − 5i 4i + 36 36 = 2π · 1 π = . 36 18 Alternatively, we may first decompose to get A B C u = + + . u + 4 n + 1 (u + 1)2 (u + 1)2 (u + 4) Here we immediately get A= −4 4 =− (−3)2 9 and C= −1 1 =− . 3 3 Then by insertion, rearrangement and reduction, B u+1 4 1 1 1 u + + (u + 1)2 (u + 4) 9 u + 4 3 (u + 1)2 1 1 9u + 4(u + 1)2 + 3(u + 4) · 9 (u + 1)2 (u + 4) 1 1 4(u + 1)2 + 12(u + 1) · 9 (u + 1)2 (u + 4) 1 4 u+1+3 4 · = · .

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