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21.4. Evaluate the following integrals in which a and b are nonzero real con- stants: 2x2 +1 dx d.x dx. x4 + 5x2 + 6 (c) (b) 6л4 + 5а? + 1 x4 +1

21.4. Evaluate the following integrals in which a and b are nonzero real con- stants: 2x2 +1 dx d.x dx. x4 + 5x2 + 6 (c) (b) 6л4 + 5а? + 1 x4 +1

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The question is about complex analysis and Calculus of Residues.

One of the 2 pictures is an example of a solution.

Please solve parts a, b and c.

Example 21.3.2. As another example, let us evaluate x sin ax dx xª + 4 where a + 0. This is the imaginary part of the integral I, = , xeias dæ/(x* +4) which, in terms of z and for the closed contour in the UHP (when a > 0), becomes zeiaz I2 = . d 2πί ) Resf (2)] z4 + 4 for a > 0, (21.6) j=1 where C is the large semicircle in the UHP. The singularities are determined by the zeros of the denominator: 24 +4 = 0 or z = 1+i,-1±i. Of these four simple poles only two, 1+i and -1+i, are in the UHP. We now calculate the residues: zeiaz Res[f(1+ i)] = lim (z – 1– i). z-1+i (z – 1– i)(z – 1 + i)(z+1 – i)(z+1+i) (1+i)eia(1+i) (2i) (2)(2+ 2i) eia e-c 8i žečaz Res[f(-1+i)] = lim (z +1– i). z--1+i (z +1– i)(z+1+ i)(z – 1 – i)(z – 1+i) (-1+i)e²a(-1+i) (2i)(-2)(-2+2i) -ia.-a 8i Substituting in Equation (21.6), we obtain I2 = 2ni- 8i - e-i«) = ie-" sin a. Thus, x sin ax dx x4 + 4 Im(I2) = 2 -a sin a for a > 0. (21.7) For a < 0, we could close the contour in the LHP. But there is an easier way of getting to the answer. We note that -a > 0, and Equation (21.7) yields a sin ax dx = x sin[(-a)x] dx sin(-a) re“ sin a. x4 + 4 x4 + 4 We can collect the two cases in x sin ax dx x4 + 4 -|a| 2 sin a.
Transcribed Image Text:Example 21.3.2. As another example, let us evaluate x sin ax dx xª + 4 where a + 0. This is the imaginary part of the integral I, = , xeias dæ/(x* +4) which, in terms of z and for the closed contour in the UHP (when a > 0), becomes zeiaz I2 = . d 2πί ) Resf (2)] z4 + 4 for a > 0, (21.6) j=1 where C is the large semicircle in the UHP. The singularities are determined by the zeros of the denominator: 24 +4 = 0 or z = 1+i,-1±i. Of these four simple poles only two, 1+i and -1+i, are in the UHP. We now calculate the residues: zeiaz Res[f(1+ i)] = lim (z – 1– i). z-1+i (z – 1– i)(z – 1 + i)(z+1 – i)(z+1+i) (1+i)eia(1+i) (2i) (2)(2+ 2i) eia e-c 8i žečaz Res[f(-1+i)] = lim (z +1– i). z--1+i (z +1– i)(z+1+ i)(z – 1 – i)(z – 1+i) (-1+i)e²a(-1+i) (2i)(-2)(-2+2i) -ia.-a 8i Substituting in Equation (21.6), we obtain I2 = 2ni- 8i - e-i«) = ie-" sin a. Thus, x sin ax dx x4 + 4 Im(I2) = 2 -a sin a for a > 0. (21.7) For a < 0, we could close the contour in the LHP. But there is an easier way of getting to the answer. We note that -a > 0, and Equation (21.7) yields a sin ax dx = x sin[(-a)x] dx sin(-a) re“ sin a. x4 + 4 x4 + 4 We can collect the two cases in x sin ax dx x4 + 4 -|a| 2 sin a.
21.4. Evaluate the following integrals in which a and b are nonzero real con- stants: (e) 2x2 + 1 dx dx (a) / dx. x4 + 5x² + 6 (b) 6x4 + 5x² + 1 x4 +1 COs x dx COS аx dx (d) . T2 + a²)² (x² + b²)' (fĐ (e) d.x. (x² + 1)²´ 1)2" (x2 + 62)2 * 2x2 dx x² dx (1)/ (k) / (g) (a2 + 1)²(x² + 2) 1 dx. x6 +1 (1) / (x² + a²)² * 0* p3 sin ax dx. x6 +1 x2 +1 dx. x2 + 4 x dx (6) / (1) (x2 + 4x + 13)² * x sin x dx (0) / x COs x dx dx (m) / (n) / 22 — 2л + 10 x² – 2x + 10 x2 + 1 x? dx dx (4) (x2 + 4)² (x² + 25) COs ax dx. x2 + b2 (p) (*) (x² + 4) *
Transcribed Image Text:21.4. Evaluate the following integrals in which a and b are nonzero real con- stants: (e) 2x2 + 1 dx dx (a) / dx. x4 + 5x² + 6 (b) 6x4 + 5x² + 1 x4 +1 COs x dx COS аx dx (d) . T2 + a²)² (x² + b²)' (fĐ (e) d.x. (x² + 1)²´ 1)2" (x2 + 62)2 * 2x2 dx x² dx (1)/ (k) / (g) (a2 + 1)²(x² + 2) 1 dx. x6 +1 (1) / (x² + a²)² * 0* p3 sin ax dx. x6 +1 x2 +1 dx. x2 + 4 x dx (6) / (1) (x2 + 4x + 13)² * x sin x dx (0) / x COs x dx dx (m) / (n) / 22 — 2л + 10 x² – 2x + 10 x2 + 1 x? dx dx (4) (x2 + 4)² (x² + 25) COs ax dx. x2 + b2 (p) (*) (x² + 4) *
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