Question

Let $A=\underset{0}{\overset{1}{\int }}\frac{e^{t}dt}{1+t}$ then $\underset{a-1}{\overset{a}{\int }}\frac{e^{-t}dt}{t-a-1}$ has the value

• A. $A{e}^{-a}$
• B. $-A{e}^{-a}$
• C. $-Aa{e}^{-1}$
• D. $A{e}^a$

Answer 1

Answer: (B)

$I=\underset{a-1}{\overset{a}{\int }}\frac{e^{-t}}{t-a-1}dt$ put $t=a-1+y$ (so that lower limit becomes zero)
$\therefore I=\underset{0}{\overset{1}{\int }}\frac{e^{1-a-y}}{y-2}dy$ (now usingking) $I=\underset{0}{\overset{1}{\int }}\frac{e^{1-a-1+y}}{1-y-2}dy=-e^{-a}\underset{0}{\overset{1}{\int }}\frac{e^y}{1+y}dy=-e^{-a}A\Rightarrow (B)$