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Complex analysis.

$\int_{0}^{\infty}\frac{sin(2x)}{x(x^2+4)}dx$

Answer 1

1 Solution het fiza iaz 2(22+4) Consider the Pintegran ffres da coleu c cn the closed . contoor consisting of the upper of the large clode (Z)=R and real axis from a&tor idented at 2=0 by a small circler of radiu or show in figure 18 r R -R Poles of fizs are given Z(2+4)=0 by

2=0, Izi fiz, has only one bole of ander 1 at z = ai cwithin The pole z=0 in aroided by identation By cechy's Gresidues theorem s frodas armi (Sum of residues coithin C) fixidx + fizedz + af ferske + Par to af fiody zari B2,62 21) 97 -R - Sinu 2 1 244) to as 12l+o Consequently, by Jordan's lemma iaz line RAS ਖੇ ) ਲ੫) dz line Roto S fezidz co

13 eidz clicy z fiz? = lim zo (22+4) Zo Therefore, liu fezidz = -i (T-0) 4 AB Ho 8 ill an are & Cosp of the carde Izlar live z fiziok (econstant) then Roto and elice s fierdz = i[B_)* Ryo AB Here negective sign is taken because the semi Colerin negatively arriented oriented Making or so, R-1 in 12), and obsowing (21, (3) s feride to + fende Ti i 4 ani. Res (2=22) -) firsdx = / iax x(2²+4) dr. Dit dui. Restz-ai)

idz Now fazla مهام ZlZ²+4) Z(2-21)(< tai) iaz e - Res(222i) Lt (2-21). Zaai Z12-21) (2+2 i) lalai) clar e = = Lt 2+2i Z(2+21) (ai laitail ē 4 Res (2=21) ē" (21)4i) Using this in (4) iar dx = T 4 tani. x(x+4) TT See ах xx²+4) dx e Tie 4 i 4 4

Equeeting imaginary parts, are 6 how sas Sinax dx= TT u Tie x(2²+4) 4 Sinan x(x²+4)