Question

${ \displaystyle\tt \int_{0}^{1} \int_{0} ^{ x_{1}} \int_{ 0 }^{x_{2}}... \int_{ 0 }^{x_{99}} ln(1 -x_{100} ) dx_{100}... dx_{1}}$​

Answer 1

$\small{ \displaystyle\tt \int_{0}^{1} \int_{0} ^{ x_{1}} \int_{ 0 }^{x_{2}}... \int_{ 0 }^{x_{99}} ln(1 -x_{100} ) dx_{100}... dx_{1}}$

$\small{ \displaystyle\tt = - \int_{0}^{1} \int_{0} ^{ x_{1}} \int_{ 0 }^{x_{2}}... \int_{ 0 }^{x_{98}} \sum_{n = 1}^{ \infty \frac{}{} \frac{}{} } \frac{1}{n} \int_{0}^{x_{99}} {x}^{n}_{100} \: dx_{100}...dx_{1} }$

$\small{ \displaystyle\tt = -\sum_{n = 1}^{ \infty \frac{}{} \frac{}{} } \frac{1}{n}\int_{0}^{1} \int_{0} ^{ x_{1}} \int_{ 0 }^{x_{2}}... \int_{ 0 }^{x_{98}} \frac{{x}^{n + 1}_{99}}{n + 1} \: dx_{99}...dx_{1} }$

$\displaystyle = \tt - \sum_{n = 1}^{ \infty } \frac{1}{n(n + 1)...(n + 101)}$

$\displaystyle \tt= - \frac{1}{(100)!_{} } \sum_{n = 1}^{ \infty } \frac{(n - 1)!(100)!}{(n + 101)!}$

$\displaystyle \tt= - \frac{1}{(100)!_{} } \sum_{n = 1}^{ \infty} \int_{0}^{1} {x}^{n - 1} (1 - x {)}^{100} \: dx$

$\displaystyle = \tt - \frac{1}{(100!)} \int_{0}^{1} \frac{(1 - x)^{100} }{1 - x} \: dx$

$\displaystyle = \tt - \frac{1}{(100)!} \int_{0}^{1} {x}^{99} \: dx$

$\huge= - \frac{1}{(100)!100}$