Question

Let $A={\int }_0^1\frac{e^t}{1+t}dt$, then ${\int }_{a-1}^a\frac{e^{-t}}{t-a-1}dt$ has the value.

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

Answer 1

The correct option is B $-A{e}^{-a}$

Given, $A{\int }_0^1\frac{e^t}{1+t}dt$

Now, $I={\int }_{a-1}^a\frac{a^{-t}}{t^{-a}-1}$

Let $x=a-t\Rightarrow dx=-dt$

$I={\int }_1^0\frac{e^{x-a}}{-x-1}(-dx)[\because x_1=a-(a-1)=1x_2=a-a=0]$

$={\int }_1^0\frac{e^{x-a}}{x+1}dx$

$=-{\int }_1^0\frac{e^{x-a}}{a+1}dx[\because {\int }_a^{b}f\left(x\right)dx=-{\int }_a^{b}f\left(x\right)dx]$

By lchanging variable property

$I=-{\int }_1^0\frac{e^{t-a}}{t+1}dt=-{\int }_1^0\frac{e^{t}dt}{t+1}{e}^{-a}$

$=-A{e}^{-a}[\because A={\int }_0^1\frac{e^{t}dt}{t+1}]$

$\therefore$ Option $B$ is correct