"I was doing a integral, the last part..." was answered by our community

Chloe Arnold

Answered question

2022-10-28

I was doing a integral, the last part is

\int_{0}^{\frac{π}{2}} x^{3} \csc x d x

Answer & Explanation

ohhappyday890b

Beginner2022-10-29Added 12 answers

Integrating by parts three times, we get

\int_{0}^{π / 2} x^{3} e^{i k x} d x = i^{k - 1} \frac{π^{3}}{8 k} + i^{k} \frac{3 π^{2}}{4 k^{2}} + i^{k + 1} \frac{3 π}{k^{3}} + \frac{6}{k^{4}} (1 - i^{k})

Therefore, using

\sin (x) = \frac{e^{i x} - e^{- i x}}{2 i}

\begin{aligned} \int_{0}^{π / 2} x^{3} \csc (x) d x \\ = \int_{0}^{π / 2} x^{3} \frac{2 i e^{- i x} d x}{1 - e^{- 2 i x}} \\ = 2 i \sum_{k = 0}^{\infty} \int_{0}^{π / 2} x^{3} e^{- (2 k + 1) i x} d x \\ (*) & = 2 i \sum_{k = 0}^{\infty} (- 1)^{k + 1} \frac{3 π^{2} i}{4 (2 k + 1)^{2}} + (- 1)^{k} \frac{6 i}{(2 k + 1)^{4}} \\ = \frac{3 π^{2}}{2} \sum_{k = 0}^{\infty} \frac{(- 1)^{k}}{(2 k + 1)^{2}} - 12 \sum_{k = 0}^{\infty} \frac{(- 1)^{k}}{(2 k + 1)^{4}} \\ = \frac{3 π^{2}}{2} G - \frac{3}{64} (ζ (4, 1 / 4) - ζ (4, 3 / 4)) \\ = \frac{3 π^{2}}{2} G - \frac{1}{128} (ψ^{(3)} (1 / 4) - ψ^{(3)} (3 / 4)) \end{aligned}

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