How to calculate the value of the integrals ∫01(arctan⁡xx)2dx, ∫01(arctan⁡xx)3dx and ∫01arctan2xln⁡xxdx ?

For the first one, ∫01(arctan⁡xx)2dx=∫0π4x2csc2xdx =−x2cot⁡x∣0π4+2∫0π4xcot⁡xdx =−π216+4∑n=1∞∫0π4xsin⁡(2nx)dx =−π216+∑n=1∞sin⁡(nπ2)n2−π2∑n=1∞cos⁡(nπ2)n =G+π4ln⁡2−π216

For the second one, ∫01arctan3xx3dx=∫0π4x3cot⁡xcsc2xdx =−12x3cot2x∣0π4+32∫0π4x2cot2xdx =−π3128−32∫0π4x2+32∫0π4x2csc2xdx =−π364+32(G+π4ln⁡2−π216) =32G−π364+3p8ln⁡2−3π232

For the third one, ∫01arctan2⁡xln⁡xxdx=−∫arctan⁡xln2⁡x1+x2dx=−∑n=0∞∑k=0n(−1)n2k+1∫01x2n+1ln2⁡xdx=−14∑n=0∞(−1)n(H2n+1−12Hn(n+1)3=14∑n=1(−1)nH2nn3−18∑n=1∞(−1)nHnn3

For the first one, ∫01(arctan⁡xx)2dx=∫0π4x2csc2xdx =−x2cot⁡x∣0π4+2∫0π4xcot⁡xdx =−π216+4∑n=1∞∫0π4xsin⁡(2nx)dx =−π216+∑n=1∞sin⁡(nπ2)n2−π2∑n=1∞cos⁡(nπ2)n =G+π4ln⁡2−π216

For the second one, ∫01arctan3xx3dx=∫0π4x3cot⁡xcsc2xdx =−12x3cot2x∣0π4+32∫0π4x2cot2xdx =−π3128−32∫0π4x2+32∫0π4x2csc2xdx =−π364+32(G+π4ln⁡2−π216) =32G−π364+3p8ln⁡2−3π232

For the third one, ∫01arctan2⁡xln⁡xxdx=−∫arctan⁡xln2⁡x1+x2dx=−∑n=0∞∑k=0n(−1)n2k+1∫01x2n+1ln2⁡xdx=−14∑n=0∞(−1)n(H2n+1−12Hn(n+1)3=14∑n=1(−1)nH2nn3−18∑n=1∞(−1)nHnn3

Explained: "How to calculate the value of the integrals..."

Joseph Krupa

Answered question

2022-01-01

How to calculate the value of the integrals

\int_{0}^{1} {(\frac{\arctan x}{x})}^{2} d x

\int_{0}^{1} {(\frac{\arctan x}{x})}^{3} d x

and

\int_{0}^{1} \frac{\arctan^{2} x \ln x}{x} d x

Answer & Explanation

Bukvald5z

Beginner2022-01-02Added 33 answers

For the first one,

\int_{0}^{1} {(\frac{\arctan x}{x})}^{2} d x = \int_{0}^{\frac{π}{4}} x^{2} \csc^{2} x d x

= - x^{2} \cot x ∣_{0}^{\frac{π}{4}} + 2 \int_{0}^{\frac{π}{4}} x \cot x d x

= - \frac{π^{2}}{16} + 4 \sum_{n = 1}^{\infty} \int_{0}^{\frac{π}{4}} x \sin (2 n x) d x

= - \frac{π^{2}}{16} + \sum_{n = 1}^{\infty} \frac{\sin (n \frac{π}{2})}{n^{2}} - \frac{π}{2} \sum_{n = 1}^{\infty} \frac{\cos (n \frac{π}{2})}{n}

= G + \frac{π}{4} \ln 2 - \frac{π^{2}}{16}

godsrvnt0706

Beginner2022-01-03Added 31 answers

For the second one,

\int_{0}^{1} \frac{\arctan^{3} x}{x^{3}} d x = \int_{0}^{\frac{π}{4}} x^{3} \cot x \csc^{2} x d x

= - \frac{1}{2} x^{3} \cot^{2} x ∣_{0}^{\frac{π}{4}} + \frac{3}{2} \int_{0}^{\frac{π}{4}} x^{2} \cot^{2} x d x

= - \frac{π^{3}}{128} - \frac{3}{2} \int_{0}^{\frac{π}{4}} x^{2} + \frac{3}{2} \int_{0}^{\frac{π}{4}} x^{2} \csc^{2} x d x

= - \frac{π^{3}}{64} + \frac{3}{2} (G + \frac{π}{4} \ln 2 - \frac{π^{2}}{16})

= \frac{3}{2} G - \frac{π^{3}}{64} + \frac{3 p}{8} \ln 2 - \frac{3 π^{2}}{32}

karton

Expert2022-01-09Added 613 answers

For the third one,
$\begin{matrix} \int_{0}^{1} \frac{\arctan^{2} x \ln x}{x} d x \\ = - \int \frac{\arctan x \ln^{2} x}{1 + x^{2}} d x \\ = - \sum_{n = 0}^{\infty} \sum_{k = 0}^{n} \frac{(- 1)^{n}}{2 k + 1} \int_{0}^{1} x^{2 n + 1} \ln^{2} x d x \\ = - \frac{1}{4} \sum_{n = 0}^{\infty} \frac{(- 1)^{n} (H_{2 n + 1} - \frac{1}{2} H_{n}}{(n + 1)^{3}} \\ = \frac{1}{4} \sum_{n = 1} \frac{(- 1)^{n} H_{2 n}}{n^{3}} - \frac{1}{8} \sum_{n = 1}^{\infty} \frac{(- 1)^{n} H_{n}}{n^{3}} \end{matrix}$

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