solve-0-pi-4-ln-1-sinx-dx-

Question Number 114196 by mathdave last updated on 17/Sep/20

s o l v e

\int_{0}^{\frac{π}{4}} \ln (1 + \sin x) d x

Commented by Dwaipayan Shikari last updated on 20/Sep/20

\int_{0}^{\frac{π}{4}} {(- 1)}^{n} \underset{n = 1}{\sum^{\infty}} \frac{{s i n}^{n} x}{n}

\sum^{\infty} {(- 1)}^{n} \int_{0}^{\frac{π}{4}} \frac{{(e^{i x} - e^{- i x})}^{n}}{2^{n} i^{n} n} d x

\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \frac{e^{i n x}}{2^{n} . i^{n} n} \int_{0}^{\frac{π}{4}} {(1 - e^{- 2 i x})}^{n}

\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \frac{1}{n} \frac{e^{i n x}}{2^{n} i^{n}} (\frac{π}{4} + \frac{n}{2 i} e^{- 2 i x} + \dots . .)

Answered by mathmax by abdo last updated on 19/Sep/20

let take a try with this integral

I = \int_{0}^{\frac{π}{4}} \ln (1 + \cos (\frac{π}{2} - x)) dx = \int_{0}^{\frac{π}{4}} \ln ({2 \cos}^{2} (\frac{π}{4} - \frac{x}{2})) dx

= \frac{π}{4} \ln (2) + 2 \int_{0}^{\frac{π}{4}} \ln (\cos (\frac{π}{4} - \frac{x}{2})) dx changement \frac{π}{4} - \frac{x}{2} = t

give \int_{0}^{\frac{π}{4}} \ln (\cos (\frac{π}{4} - \frac{x}{2})) dx = \int_{\frac{π}{4}}^{\frac{π}{8}} \ln (cost) (- 2 dt)

= 2 \int_{\frac{π}{8}}^{\frac{π}{4}} \ln (cost) dt = 2 {{[t \ln (cost)]}_{\frac{π}{8}}^{\frac{π}{4}} - \int_{\frac{π}{8}}^{\frac{π}{4}} t \times \frac{- sint}{cost} dt}

= 2 {\frac{π}{4} \ln (\frac{1}{\sqrt{2}}) - \frac{π}{8} \ln (\frac{\sqrt{2 + \sqrt{2}}}{2})} + 2 \int_{\frac{π}{8}}^{\frac{π}{4}} t tant dt

\int_{\frac{π}{8}}^{\frac{π}{4}} t tant dt =_{tant = u} \int_{\sqrt{2} - 1}^{1} \frac{u arctanu}{1 + u^{2}} du

= {[\frac{1}{2} \ln (1 + u^{2}) arctanu]}_{\sqrt{2} - 1}^{1} - \int_{\sqrt{2} - 1}^{1} \frac{1}{2} \ln (1 + u^{2}) \times \frac{du}{1 + u^{2}}

= \frac{1}{2} \ln (2) \frac{π}{4} - \frac{1}{2} \ln (4 - 2 \sqrt{2}) \times \frac{π}{8} - \frac{1}{2} \int_{\sqrt{2} - 1}^{1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du

but \int_{\sqrt{2} - 1}^{1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du = \int_{0}^{1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du - \int_{0}^{\sqrt{2} - 1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du

\int_{0}^{1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du =_{u = \tan θ} \int_{0}^{\frac{π}{4}} \frac{\ln (\frac{1}{\cos^{2} θ})}{1 + \tan^{2} θ} (1 + \tan^{2} θ) d θ

= - 2 \int_{0}^{\frac{π}{4}} \ln (\cos θ) (the value of this integral is known see the

platform i dont remember its value)

\int_{0}^{\sqrt{2} - 1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du = \int_{0}^{\sqrt{2} - 1} \ln (1 + u^{2}) \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{2 n} du

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\sqrt{2} - 1} u^{2 n} \ln (1 + u^{2}) du \dots . be continued \dots