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Question Number 126986 by mnjuly1970 last updated on 25/Dec/20

\dots n i c e c a l c u l u s \dots

p r o v e t h a t ::

I := \int_{0}^{\frac{π}{2}} \frac{{c o t (x)}}{c o t (x)} d x = \frac{1}{2} (π - l n (\frac{s i n h (π)}{π}))

{x} i s f r a c t i o n a l p a r t o f x . .

Answered by Olaf last updated on 26/Dec/20

I = \int_{0}^{\frac{π}{2}} \frac{{\cot x}}{\cot x} d x

Let u = \cot x

I = \int_{\infty}^{0} \frac{{u}}{u} . \frac{d u}{(- 1 - u^{2})}

I = \int_{0}^{\infty} \frac{{u}}{u (1 + u^{2})} d u

I = \underset{n = 0}{\sum^{\infty}} \int_{n}^{n + 1} \frac{{u}}{u (1 + u^{2})} d u

I = \underset{n = 0}{\sum^{\infty}} \int_{n}^{n + 1} \frac{u - n}{u (1 + u^{2})} d u

I = \underset{n = 0}{\sum^{\infty}} \int_{n}^{n + 1} [\frac{n u + 1}{u^{2} + 1} - \frac{n}{u}] d u

I = \underset{n = 0}{\sum^{\infty}} {[\frac{n}{2} \ln (1 + u^{2}) - \arctan u - n \ln u]}_{n}^{n + 1}

\dots t o b e c o n t i n u e d \dots

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