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Q.
Let $A=\int\limits_0^1 \frac{e^t d t}{1+t}$ then $\int\limits_{a-1}^a \frac{e^{-t} d t}{t-a-1}$ has the value
Integrals
Solution:
$I=\int\limits_{ a -1}^{ a } \frac{ e ^{- t }}{ t - a -1} dt $ put $t = a -1+ y$ (so that lower limit becomes zero)
$\therefore I =\int\limits_0^1 \frac{ e ^{1- a - y }}{ y -2} dy $ (now usingking) $I=\int\limits_0^1 \frac{e^{1-a-1+y}}{1-y-2} d y=-e^{-a} \int\limits_0^1 \frac{e^y}{1+y} d y=-e^{-a} A \Rightarrow(B)$