Question-107328

Question Number 107328 by mathdave last updated on 10/Aug/20

Commented by mnjuly1970 last updated on 10/Aug/20

Commented by mnjuly1970 last updated on 10/Aug/20

Commented by mathdave last updated on 10/Aug/20

p l s h w u g o t a l l d s

Commented by mathdave last updated on 10/Aug/20

d s i s n t a c t u a l l y s o l v e b y u

Commented by mnjuly1970 last updated on 10/Aug/20

p l e a s e : n o w, s o l v e i t

i n y o u r o w n w a y!

Commented by mathmax by abdo last updated on 10/Aug/20

can you show how you get the value of Σ \frac{{(- 1)}^{n}}{2 n + 1} \ln (2 n + 1) ?

Answered by mathmax by abdo last updated on 10/Aug/20

I = \int_{0}^{\infty} \frac{lnx}{e^{x} + e^{- x}} dx \Rightarrow I = \int_{0}^{\infty} \frac{e^{- x} \ln (x)}{1 + e^{- 2 x}} dx

= \int_{0}^{\infty} e^{- x} \ln (x) \sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- 2 nx} dx

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} e^{- (2 n + 1) x} lnx dx changement (2 n + 1) x = t give

\int_{0}^{\infty} e^{- (2 n + 1) x} \ln (x) dx = \int_{0}^{\infty} e^{- t} \ln (\frac{t}{2 n + 1}) \frac{dt}{2 n + 1}

= \frac{1}{2 n + 1} \int_{0}^{\infty} e^{- t} {\ln (t) - \ln (2 n + 1)} dt

= \frac{1}{2 n + 1} \int_{0}^{\infty} e^{- t} \ln (t) dt - \frac{\ln (2 n + 1)}{2 n + 1} \int_{0}^{\infty} e^{- t} dt

= - \frac{γ}{2 n + 1} - \frac{\ln (2 n + 1)}{2 n + 1} \Rightarrow I = - γ \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} \ln (2 n + 1)}{2 n + 1}

= - \frac{π γ}{4} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} \ln (2 n + 1) rest to find this sum \dots be continued . .

Commented by mnjuly1970 last updated on 10/Aug/20

p e r f e c t . m e r c e y

Commented by mathdave last updated on 10/Aug/20

u h a d r e a l l y d o n e w e l l o o o w e l l d o n e s i r

Commented by mathmax by abdo last updated on 10/Aug/20

you are welcome