Evaluate the following definite integral using the method of Example 10.2 (page 67): 90 0 x...

 

  

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Evaluate the following definite integral using the method of Example 10.2 (page 67): 90 0 x² +1 dx for a real with a > 0. Hint: Since the integrand is an even function of x you may use * cos(ax) dx cos (ax) - Pj x² +1 -dx x² +1 where P denotes the Cauchy principal value (as in lecture notes, Eq. (10.4)). For a>0 you may close the integration contour -RR in the upper complex half-plane. To show that the contour integral over the half-circle CR with radius R vanishes for R→∞, you may use again 2² +12R-1 for ze C, and in addition |ea|=e" ay≤1 for z = C, where y = Im (z). Ex. 10.2. Compute xsinx x²+2x+2 Let dx. z Z f(2)= 22 +22+2 (z-21)(z- zj) where 2₁-1+1. Note: z₁ is a simple pole of f(z)elz in the upper-half plane with residue b₁ - Res f(z)elz= 21 elz 21-2₁ Use the contour C= Lg + CR shown in Fig. 10.2. We see xelx dx=2rib₁-f(z)e¹² dz +2x+2 R want to bound this Im z CR R Figure 10.2: Contour for Ex. 10.2 Rez (10.16) (10.17) (10.18) (10.19) Note: If(z)| ≤ Mg where MR = R/(R-√2)2 and le=ey <1 so 20¹² des TR² (R-√2)2 but this does not go to zero as R➡00. We need to be more careful: MRTR = (z)ez dz=f(Re)eRe Rel® de. Now if(Re)s Mg and le Re Ise-Rsine so (2)e¹² dzs M&R Se Thus, Sesine -Reine de 52 2 -2R0/1 = 2 (1-6) 1/2 e < e-Rsine de. We use Jordan's Inequality to bound the Integral: since sin 8> 20/ for 0<8<n/2 (see Fig. 10.3), and therefore f Sc₁"(z)6¹" dz\ < M₂R = TMR Oas R-00 xsinx x2+2x+2 dx = Im(2x/b₁) = (sin1 + cos 1). sing y-28/m n/2 11 (10.20) Figure 10.3: Jordan's Inequality (10.21) (10.22) (10.23a) (10.23b) (10.24a) (10.24b) (10.25) Evaluate the following definite integral using the method of Example 10.2 (page 67): 90 0 x² +1 dx for a real with a > 0. Hint: Since the integrand is an even function of x you may use * cos(ax) dx cos (ax) - Pj x² +1 -dx x² +1 where P denotes the Cauchy principal value (as in lecture notes, Eq. (10.4)). For a>0 you may close the integration contour -RR in the upper complex half-plane. To show that the contour integral over the half-circle CR with radius R vanishes for R→∞, you may use again 2² +12R-1 for ze C, and in addition |ea|=e" ay≤1 for z = C, where y = Im (z). Ex. 10.2. Compute xsinx x²+2x+2 Let dx. z Z f(2)= 22 +22+2 (z-21)(z- zj) where 2₁-1+1. Note: z₁ is a simple pole of f(z)elz in the upper-half plane with residue b₁ - Res f(z)elz= 21 elz 21-2₁ Use the contour C= Lg + CR shown in Fig. 10.2. We see xelx dx=2rib₁-f(z)e¹² dz +2x+2 R want to bound this Im z CR R Figure 10.2: Contour for Ex. 10.2 Rez (10.16) (10.17) (10.18) (10.19) Note: If(z)| ≤ Mg where MR = R/(R-√2)2 and le=ey <1 so 20¹² des TR² (R-√2)2 but this does not go to zero as R➡00. We need to be more careful: MRTR = (z)ez dz=f(Re)eRe Rel® de. Now if(Re)s Mg and le Re Ise-Rsine so (2)e¹² dzs M&R Se Thus, Sesine -Reine de 52 2 -2R0/1 = 2 (1-6) 1/2 e < e-Rsine de. We use Jordan's Inequality to bound the Integral: since sin 8> 20/ for 0<8<n/2 (see Fig. 10.3), and therefore f Sc₁"(z)6¹" dz\ < M₂R = TMR Oas R-00 xsinx x2+2x+2 dx = Im(2x/b₁) = (sin1 + cos 1). sing y-28/m n/2 11 (10.20) Figure 10.3: Jordan's Inequality (10.21) (10.22) (10.23a) (10.23b) (10.24a) (10.24b) (10.25)

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To evaluate the definite integral 0 to cosax x 1 dx using the method of Example 102 we can use the C... View the full answer