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# solut5 - Solutions to homework 5 II.31(a Consider the...

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Solutions to homework # 5 II.31. (a) Consider the rectangular contour C R consisting of the intervals I R :=[ - R, R ], S + R :=[ R, R + 2 π i /b ], J R :=[ R + 2 π i /b, - R + 2 π i /b ], S - R :=[ - R + 2 π i /b, - R ]. First note that Z J R e az d z 1 + e bz = - Z I R e a ( x +2 π i /b ) d x 1 + e bx = - e 2 π i a/b Z I R e ax d x 1 + e bx . Next, Z S + R e az d z 1 + e bz 2 π b · e Re aR e bR - 1 0 as R → ∞ , Z S - R e az d z 1 + e bz 2 π b · e - Re aR 1 - e - bR 0 as R → ∞ . So, Z -∞ e ax d x 1 + e bx = lim R →∞ 1 1 - e 2 π i a/b Z I R + Z J R e az d z 1 + e bz = 1 1 - e 2 π i a/b lim R →∞ Z C R e az d z 1 + e bz . By the Residue Theorem, the last integral is equal to 2 π i times the sum of all residues inside the contour. Since the only singularity of the integrand inside the contour is a simple pole at the point π i /b , we use the formula Res( f/g )( z 0 ) = lim z z 0 f ( z ) /g 0 ( z ) to get Z -∞ e ax d x 1 + e bx = 2 π i 1 - e 2 π i a/b lim z π i /b e az be bz = 2 π i 1 - e 2 π i a/b - e π i a/b b = 2 π i b ( e π i a/b - e - π i a/b ) = 2 π i b · 2i sin( aπ/b ) = π b sin( aπ/b ) . (b) Consider the rectangular contour C R consisting of the intervals I R :=[ - R, R ], S + R :=[ R, R + π i /a ], J R :=[ R + π i /a, - R + π i /a ], S - R :=[ - R + π i /a, - R ]. First note that Z J R sinh az d z sinh 4 az = - Z I R sinh( a ( x + π i /a )) d x sinh(4

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