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Sa se calculeze

\( \lim_{n\to\infty}\int_0^n ln(1+e^{-x})dx \).

Folosim dezvoltarea in serie \( ln(1+x)=\sum\limits_{n\geq0}(-1)^n\frac{x^{n+1}}{n+1} \).
\( \Rightarrow \int\limits_0^n ln(1+e^{-x})=
\int\limits_0^n\sum\limits_{k\geq0}(-1)^k\frac{e^{-x(k+1)}}{k+1}= \)
\( =\sum\limits_{k\geq0}\frac{(-1)^k}{k+1}\int\limits_0^n e^{-x(k+1)}= \)
\( =\sum\limits_{k\geq0}\frac{(-1)^{k+1}}{(k+1)^2}(e^{-n(k+1)}-1)= \)
\( =e^{-n}\sum\limits_{k\geq0}\frac{(-1)^{k+1}}{(k+1)^2}e^{-nk}-\sum\limits_{k\geq0}\frac{(-1)^k}{(k+1)^2}= \)
\( =e^{-n}C+\frac{\pi^2}{12}\rightarrow\frac{\pi^2}{12} \)

Cezar Lupu wrote:Sa se calculeze

\( \lim_{n\to\infty}\int_0^n ln(1+e^{-x})dx \).

A double integration trick yields the answer too.

K!! wrote:

Cezar Lupu wrote:Sa se calculeze

\( \lim_{n\to\infty}\int_0^n ln(1+e^{-x})dx \).

A double integration trick yields the answer too.

Can you be more specific? My approach is the same as Ciprian's above.

\( \ln \left( {1 + e^{ - x} } \right) = \int_0^{\exp ( - x)} {\frac{1}
{{1 + u}}\,du} . \)

By reversing integration order we get a single integral which can be solved using the geometric series, the rest follows by using the Basel problem.