# Math Help - Today's calculation of integral #4

1. ## Today's calculation of integral #4

No matter what $k$ is. Let $
\alpha=\int_0^1\frac{e^u}{1+u}\,du
$
& $
\beta=\int_{k-1}^k\frac{e^{-x}}{x-k-1}\,dx.
$
Assume that $
\lambda\cdot \alpha=\beta.
$
Compute $\lambda.$

2. Originally Posted by Krizalid
No matter what $k$ is. Let $
\alpha=\int_0^1\frac{e^u}{1+u}\,du
$
& $
\beta=\int_{k-1}^k\frac{e^{-x}}{x-k-1}\,dx.
$
Assume that $
\lambda\cdot \alpha=\beta.
$
Compute $\lambda.$
let $-x=u-k \mbox{ then } dx=-du$

$
\beta=\int_{k-1}^k\frac{e^{-x}}{x-k-1}\,dx=\int_{1}^{0}\frac{e^{u-k}}{-u-1}(-du)=\frac{1}{e^k}\int_{0}^{1}\frac{e^u}{u+1}du
$

so $\lambda =\frac{\beta}{\alpha}=\frac{\frac{1}{e^k}\int_{0}^ {1}\frac{e^u}{u+1}du}{\int_{0}^{1}\frac{e^u}{u+1}d u}=\frac{1}{e^k}$

3. That's correct.