13oh...thanx bhaaiya...
b=\int_{0}^{1}{\frac{e^t}{t+1}} dt
then\int_{a-1}^{a}{\frac{e^-t}{t-1-a}}dt=?
in terms of a and b
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6 Answers
62substitute z=a-t
z=a-t \Rightarrow dz= -dt\\ \\ I=\int_{1}^{0}{\frac{e^{z-a}}{-z-1}}dz\\ I=-e^{-a}\int_{1}^{0}{\frac{e^z}{z+1}}dz\\ I=e^{-a}\int_{0}^{1}{\frac{e^z}{z+1}}dz\\ I = e^{-a}b
62initially t went from a-1 to a
now z=a-t
so limit will go from (a-a+1) to (a-a) = 1 to 0