Question

R R 5. To compute 1 = lim 2 COS dr and J = lim 22+1 sinc dx simultaneously .22 +1 R R0 R R using Residue Theorem, let f(x) 22 +1

Answer 1

Solution R Given ; doc I= lim Ryoo x Cosx x² + 1 R Labo dx ac Sink Lim Rtoo xc² + 1 -R zeiz 1) Take f(z) : z2 + 1 vo where, I= x + c y Take of yao, than or Rz = x f (R- x) = x² + de eise ( cos x + c $in x) Я x²+1 So, R f (R2) = x Coss x²+1 and I f (R₂) = x Sins x²+1 2) f (2) = zeiz z²+1 zeiz (z - i)(2+i) two pole of order 1 Zoi, 40 hos

And, Res (f(2), i) = Lim (2-2) f (2) zyi alim (2-1) zeiz zi z²+1 ze zeiz zti Zti i ciri iti elo or Res (&(2), i) = a ૨ ૯ 3) Again, ze z2+1 dz Z 1 $ 4(2) dz reiz = ani. Res (f (2), :) - 2ri x F C Ś P(z) dz x Casse dx fillin Also, se Sinse dx x²+1 Ryoo -R i e of (2) dz I ti J T . = I ti] Hence, T I=0 & J =

R Given, x Cosse doc H 11 lim R700 x² + 1 -R take f (x) = ac Cosse 20²+1 So, f(-x) = (-x) Cos (-x) (-x)² + 1 x Cos (x) x² + 1 20 i 2.0 f(sc) & Therefore. it an odd for odd function. fonction , And know that 99 f(x) dx = 0 R So, x Cosx x²+1 dx Ryoo -R