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					 The value of the integral
∞∫-∞ sin x dx is x² + 2x + 2 
evaluated using contour integration and the residue theorem is 
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-  
πsin (1) e  -  
- πcos (1) e  -  
sin (1) e  -  
cos (1) e  
 -  
 
Correct Option: A
| I = ∞∫-∞ | ||
| x² + 2x + 2 | 
| let f(x) = | ||
| z² + 2z + 2 | 
Then poles of f(z) are given by z² + 2z + 2 = 0
∴ z = – 1 ± i
R1 = Res: (f(z): z = -1 + i)
| Ltz→-1±1[z-(-1+i)] | ||
| [z-(-1+1)][z-(-1-i)] | 
| = | = | ||
| -1 + i + 1 + i | 2i | 

| = IM[πe-1(cos(1) - isin(1))] = - | ||
| e |