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I believe I have evaluated the following integral, and was wondering if I can get a check to my work from my fellow MSE users please! Here is my work and the integral.

$$I:=\int_{0}^{t}\frac{\arctan(x)}{1-x^2}\,dx$$

$$I=\arctan(t)\mathrm{arctanh}(t)-\frac{1}{2}\int_{0}^{t}\frac{\log(\frac{1+x}{1-x})}{1+x^2}\,dx$$

Let $$x\longrightarrow{\frac{1-x}{1 +x}}$$

$$I=\arctan(t)\mathrm{arctanh}(t)-\frac{1}{2}\int_{1}^{\frac{1-t}{1+t}}\frac{\log(x)}{1+x^2}\,dx$$

$$I=\arctan(t)\mathrm{arctanh}(t)-\frac{1}{2}I_{1}$$

Now consider the integral:

$$I_{1}:=\int_{1}^{\frac{1-t}{1+t}}\frac{\log(x)}{1+x^2}\,dx$$

$$I_{1}=\int_{1}^{\frac{1-t}{1+t}}\log(x)\sum_{n=0}^{\infty}(-1)^nx^{2n}\,dx$$

$$I_{1}=\sum_{n=0}^{\infty}(-1)^n\int_{1}^{\frac{1-t}{1+t}}x^{2n}\log(x)\,dx$$

$$I_{1}=\sum_{n=0}^{\infty}(-1)^n[\frac{(\frac{1-t}{1+t})^{2n+1}}{2n+1}\log(\frac{1-t}{1+t})-\frac{(\frac{1-t}{1+t})^{2n+1}}{(2n+1)^2}+\frac{1}{(2n+1)^2}]$$

$$I_{1}=\log(\frac{1-t}{1+t})\arctan(\frac{1-t}{1+t})-\mathrm{Ti}_{2}(\frac{1-t}{1+t})+G$$

Putting everything together we get the resut:

$$I=\mathrm{arctanh}(t)[\arctan(t)+\arctan(\frac{1-t}{1+t})]-\frac{1}{2}G+\frac{1}{2}\mathrm{Ti}_{2}(\frac{1-t}{1+t})$$

Where $\mathrm{Ti}_{2}(x)$ denotes the inverse tangent integral, $G$ denotes Catalan’s constant.

Is this right? Can it be simplified? Anyone got cool and related integrals?

Jean-Claude Arbaut
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Person
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  • wolfram evaluates the indefinite integral in terms of Li_2 : https://www.wolframalpha.com/input?i2d=true&i=integral%5C%2840%29+Divide%5Barctan%5C%2840%29x%5C%2841%29%2C1-Power%5Bx%2C2%5D%5D%5C%2841%29 – coffeemath Jul 17 '23 at 00:53
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    Isn't there a sign error in the change of variable $x\longrightarrow{\frac{1-x}{1 +x}}$? You should get $\log(1/x)=-\log(x)$ instead of $\log(x)$. – Jean-Claude Arbaut Jul 17 '23 at 01:04
  • Numerically speaking, your conjectured result looks correct as long as $t \in (-1,1)$. – Accelerator Jul 17 '23 at 03:16
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    To simplify a bit more $$\arctan(t)+\arctan\left(\frac{1-t}{1+t}\right)=\frac \pi 4$$ for $0 \leq t \leq 1$ – Claude Leibovici Jul 17 '23 at 08:15
  • @ClaudeLeibovici It's even true for all $t>-1$. For $t<-1$ it's $-3\pi/4$. The LHS is continuous and has null derivative, hence is constant on every connected set where it's defined, and it's only undefined at $-1$. – Jean-Claude Arbaut Jul 17 '23 at 09:01
  • Note that to denote the inverse of hyperbolic functions, we do not use 'arc', but rather 'ar' $$\operatorname{artanh}(x)$$ $$\operatorname{arsinh}(x)$$ $$\operatorname{arsech}(x)$$ – phi-rate Jul 17 '23 at 21:21
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    @phi-rate Highly depends on who is "we". In France I have always seen them written argth,argsh and argch, with the direct ones th, sh and ch. It seems to me arctanh, atanh or $\mathrm{tanh}^{−1}$ are also fairly common in the anglo-saxon world. However, I didn't check but ISO/CEI 80000-2 is apparently recommending tanh and artanh. But see also https://math.meta.stackexchange.com/questions/9668/what-is-the-accepted-notation-for-natural-logarithms-trig-functions-and-powers – Jean-Claude Arbaut Jul 18 '23 at 11:49

1 Answers1

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This is somewhat similar to your work. The only major differences are that I used IBP twice and Taylor-expanded $\arctan x$.

$$ \begin{align} I &:= \int_{0}^{t}\frac{\arctan x}{1-x^{2}}dx \\ &= \arctan\left(t\right)\operatorname{artanh}\left(t\right)-\frac{1}{2}\int_{0}^{t}\ln\left(\frac{1+x}{1-x}\right)\frac{1}{1+x^{2}}dx \\ &= \arctan\left(t\right)\operatorname{artanh}\left(t\right)-\frac{1}{2}\int_{1}^{\frac{1-t}{1+t}}\frac{\ln x}{1+x^{2}}dx \\ &= \arctan\left(t\right)\operatorname{artanh}\left(t\right)-\frac{1}{2}\arctan\left(\frac{1-t}{1+t}\right)\ln\left(\frac{1-t}{1+t}\right)+\frac{1}{2}\int_{1}^{\frac{1-t}{1+t}}\frac{\arctan x}{x}dx \\ &= \operatorname{artanh}\left(t\right)\left(\arctan t+\arctan\left(\frac{1-t}{1+t}\right)\right)+\frac{1}{2}\int_{0}^{\frac{1-t}{1+t}}\frac{\arctan x}{x}dx-\frac{1}{2}\int_{0}^{1}\frac{\arctan x}{x}dx \\ &= \operatorname{artanh}\left(t\right)\left(\arctan t+\arctan\left(\frac{1-t}{1+t}\right)\right)+\frac{1}{2}\operatorname{Ti}_{2}\left(\frac{1-t}{1+t}\right)-\frac{1}{2}\int_{0}^{1}\frac{1}{x}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{2n+1}}{2n+1}dx \\ &= \operatorname{artanh}\left(t\right)\left(\arctan t+\arctan\left(\frac{1-t}{1+t}\right)\right)+\frac{1}{2}\operatorname{Ti}_{2}\left(\frac{1-t}{1+t}\right)-\frac{1}{2}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{2n+1}\int_{0}^{1}x^{2n}dx \\ &= \operatorname{artanh}\left(t\right)\left(\arctan t+\arctan\left(\frac{1-t}{1+t}\right)\right)+\frac{1}{2}\operatorname{Ti}_{2}\left(\frac{1-t}{1+t}\right)-\frac{1}{2}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(2n+1\right)^{2}}dx \\ &= \operatorname{artanh}\left(t\right)\left(\arctan t+\arctan\left(\frac{1-t}{1+t}\right)\right)+\frac{1}{2}\operatorname{Ti}_{2}\left(\frac{1-t}{1+t}\right)-\frac{G}{2} \\ \end{align} $$

Accelerator
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