Find-the-value-of-the-folloing-integral-determinant-0-2-1-1-cosx-1-3-dx-

Question Number 208176 by mnjuly1970 last updated on 07/Jun/24

F i n d t h e v a l u e o f t h e

f o l l o i n g i n t e g r a l .

\begin{matrix} Ω = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \sqrt[3]{c o s x}} d x = ? \end{matrix}

Commented by Frix last updated on 07/Jun/24

See question 208140

Answered by Berbere last updated on 07/Jun/24

\frac{1}{1 + \sqrt[3]{c o s (x)}} = Σ {(- 1)}^{n} {c o s}^{\frac{n}{3}} (x)

Ω = \sum_{n ⩾ 0} \int_{0}^{\frac{π}{2}} {(- 1)}^{n} {c o s}^{2 (\frac{n}{6} + \frac{1}{2}) - 1} (x) {s i n}^{2 (\frac{1}{2}) - 1} (x) d x

= \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{2} β (\frac{n}{6} + \frac{1}{2}, \frac{1}{2}) = \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{2} \frac{Γ (\frac{n}{6} + \frac{1}{2}) Γ (\frac{1}{2})}{Γ (\frac{n}{6} + 1)}

N = \underset{k \in {0, 1, 2, 3, 4, 5)}{\cup} (6 N + k)

= \underset{k = 0}{\sum^{5}} \underset{m = 0}{\sum^{\infty}} \frac{{(- 1)}^{6 m + k}}{2} . \frac{Γ (m + \frac{k}{6} + \frac{1}{2})}{Γ (m + \frac{k}{6} + 1)} Γ (\frac{1}{2})

= Γ (\frac{1}{2}) \underset{k = 0}{\sum^{5}} \frac{{(- 1)}^{k}}{2} \sum_{m ⩾ 0} \frac{\frac{Γ (m + \frac{k}{6} + \frac{1}{2})}{Γ (m)}}{\frac{Γ (m + \frac{k}{6} + 1)}{Γ (m)}} . Γ (m + 1) . \frac{1^{m}}{m!}

= \frac{\sqrt{π}}{2} \underset{k = 0}{\sum^{5}}_{2} F_{1} (\frac{k}{6} + \frac{1}{2}, 1; \frac{k}{6} + 1; [1])

Commented by Frix last updated on 07/Jun/24

Nice!

Commented by Berbere last updated on 10/Jun/24

t h a n k Y o u s i r H a v e a n i c e D a y

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