Tham khảo tài liệu 'advanced mathematical methods for scientists and engineers episode 2 part 10', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Solution Convergence. If xa f x xa as x 0 for some a 1 then the integral i xaf x dx Jo will converge absolutely. If xaf x x as x -TO for some fi 1 then the integral j xf x will converge absolutely. These are sufficient conditions for the absolute convergence of tt xaf x dx. Jo Contour Integration. We put a branch cut on the positive real axis and choose 0 arg z 2n. We consider the integral of zaf z on the contour in Figure . Let the singularities of f z occur at z1 . zn. By the residue theorem i zaf z dz i2n V Res zaf z Zk . Jc k i On the circle of radius e the integrand is o e-1 . Since the length of C is 2ne the integral on C vanishes as e 0. On the circle of radius R the integrand is o R-1 . Since the length of CR is 2nR the integral on CR vanishes as R TO. The value of the integrand below the branch cut z x e 2 is f x e12n xa e12na f x In the limit as e 0 and R TO we have i xaf x dx i xa e12na f x dx i2n Res zaf z zk . Jo J-rc k i 734 . . . i2n . xf x dx 1 - J2n 2 Res zf z zk 1 - k 1 Solution In the interval of uniform convergence of th integral we can differentiate the formula i2n J x f x dx 1 - J2na j2Res z f z zk with respect to a to obtain xaf x log x dx J i2n 4n2a e12na 1 - eLa E Res zaf z log z zk - 1 _ ela 2 E Res zaf z zk i2n . . . n2a . . . . I x f x log x dx 1 - ei2na 2 Res z f z log z zk sin2 na I2Res z f z zfc k 1 k 1 Differentiating the solution of Exercise m times with respect to a yields dm i2n Jo x f x log x dx m 1 - eLa z Res z f z zk j Solution Taking the limit as a 0 G Z in the solution of Exercise yields Res zaf z Zk 1 el2na J f x dx i2n lim The numerator vanishes because the sum of all residues of zn f z is zero. Thus we can use L Hospital s rule. r f x dx lim Ves zzf z log z z f 7 -i2n e12na .
