How to evaluate $\int_{0}^1\frac{\arctan(x)}{x}\,dx$?

Question

I tried to evaluate using by-parts and some manipulation but my answer was $\frac {\pi} {12}+\frac 23 $ which is approximately 0.928466. However when I checked wolfram alpha it gave the answer as 0.915966 and I suspect I have made some mistake.

Answer 1

For any $x\in(-1,1)$ we have $$ \arctan(x) = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots\tag{1} $$ hence by dividing both sides by $x$ and apply termwise integration on $(0,1)$ we get: $$ \int_{0}^{1}\frac{\arctan x}{x}\,dx = 1-\frac{1}{9}+\frac{1}{25}-\frac{1}{49}+\ldots = K\tag{2}$$ namely Catalan's constant.

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Late addendum. Decent rational approximations for $K$ can be produced by creative telescoping. By coupling consecutive terms we have $K=\sum_{n\geq 0}\frac{8(2n+1)}{(4n+1)^2(4n+3)^2}$, and $\frac{8(2n+1)}{(4n+1)^2(4n+3)^2}$ and $\frac{1}{32n^2+6}-\frac{1}{32(n+1)^2+6}$ have the same asymptotic expansion up to the $O\left(\frac{1}{n^6}\right)$ term. This leads to $$ K \approx \frac{8}{9}+\frac{1}{38} $$ where the absolute error is less than $8\cdot 10^{-4}$.

Answer 2

Assume $$I = \int^{1}_{0}\frac{\tan^{-1}(x)}{x}dx = \tan^{-1}(x)\ln(x)\bigg|^{1}_{0}-\int^{1}_{0}\frac{\ln(x)}{1+x^2}dx$$

So $$I = -\int^{1}_{0}\sum^{\infty}_{n=0}(-1)^nx^{2n}\ln xdx = \sum^{\infty}_{n=0}(-1)^{n+1}\int^{1}_{0}x^{2n}\ln xdx$$

using by parts

$$I = \sum^{\infty}_{n=0}(-1)^{n+1}\bigg[\ln x \frac{x^{2n+1}}{2n+1}\bigg|^{1}_{0}-\frac{1}{(2n+1)}\int^{1}_{0}x^{2n}dx\bigg]$$

So $$I = \sum^{\infty}_{n=0}(-1)^{n+2}\frac{1}{(2n+1)^2} = \sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)^2}$$