3

I need to evaluate the following indefinite integral for some other definite integral $$\int\frac{x\arctan x}{x^4+1}dx$$ I found that $$\int_o^\infty\arctan{(e^{-x})}\arctan{(e^{-2x})}dx=\frac{\pi G}{4}-2I(1)$$ where $$I(t)=\int_0^1\frac{x\operatorname{Ti}_2(tx)}{x^4+1}dx$$ taking derivatives $$tI'(t)=\int_0^1\frac{x\arctan(tx)}{x^4+1}dx$$ $$(tI')'=\int_0^1\frac{x^2}{(t^2x^2+1)(x^4+1)}dx$$ The answer for the integral is the integrand in my indefinite integral plus some stuff I can integrate on my own, which is where my problem comes from. In reality, I need $$\int\frac{1}{t}\left(\int\frac{t\arctan{t}}{t^4+1}dt\right)dt$$ But it's fine if I just get an answer for the inside integral. Back to the indefinite integral. I wanted to use partial fractions $$\int\frac{x\arctan x}{(x-e^{\pi i/4})(x-e^{3\pi i/4})(x-e^{5\pi i/4})(x-e^{7\pi i/4})}dx$$ We would get four integrals of the form $\int\frac{\arctan{x}}{x+c}dx$. Using the logarithmic definition of arctangent, we get 8 integrals of the form $\int\frac{\ln{(a+bx)}}{x+c}$. That's 8 dilogarithms. I need this for another calculation, and having 8 dilogarithms is too numerous to be practical to work with. I want another way to evaluate this integral.

WAs answer

TO CLARIFY

I am fine with dilogarithms

phi-rate
  • 2,345
  • 2
    What was the original integral you wanted to work with? – Ninad Munshi Dec 17 '22 at 21:43
  • 1
    The indefinite integral thar you are trying to calculate does not a closed form in terms of elementary functions. Go back to your original integral and you can add it to your current publication, perhaps it is the focus of the first integral. More context may be needed to get a more concise answer. – A. P. Dec 17 '22 at 22:59
  • @A.P. I am fine with special functions. – phi-rate Dec 17 '22 at 23:13
  • @NinadMunshi I edited my question, and I hope it is to your satisfaction – phi-rate Dec 17 '22 at 23:13

1 Answers1

2

I am not sure that you could avoid dilogarithms except if you accept an infinite summation of Gaussian hypergeometric functions. $$I=\int \frac{x }{x^4+1}\tan ^{-1}(x)\,dx$$ Using integration by parts $$I=\frac{1}{2} \tan ^{-1}(x) \tan ^{-1}\left(x^2\right)-\frac 12\int \frac{\tan ^{-1}\left(x^2\right)}{x^2+1}\,dx$$ looks a bit more pleasant.

Using $$\tan ^{-1}\left(x^2\right)=\sum_{n=0}^\infty\frac{(-1)^n}{2 n+1}x^{4n+2}$$

$$J=\int \frac{\tan ^{-1}\left(x^2\right)}{x^2+1}\,dx$$ $$J=\sum_{n=0}^\infty (-1)^n\,\frac{x^{4 n+3}}{(2n+1)(4 n+3)}\,\,\, _2F_1\left(1,\frac{4n+3}{2};\frac{4n+5}{2};-x^2\right)$$ Notice that

$$x^{4 n+3}\, _2F_1\left(1,\frac{4n+3}{2};\frac{4n+5}{2};-x^2\right)=-(4n+3)\tan ^{-1}(x)+P_{4n+1}(x)$$ where the first polynomials are $$\left( \begin{array}{cc} 0 & 3 x \\ 1 & \frac{7 x^5}{5}-\frac{7 x^3}{3}+7 x \\ 2 & \frac{11 x^9}{9}-\frac{11 x^7}{7}+\frac{11 x^5}{5}-\frac{11 x^3}{3}+11 x \\ 3 & \frac{15 x^{13}}{13}-\frac{15 x^{11}}{11}+\frac{5 x^9}{3}-\frac{15 x^7}{7}+3 x^5-5 x^3+15 x \\ 4 & \frac{19 x^{17}}{17}-\frac{19 x^{15}}{15}+\frac{19 x^{13}}{13}-\frac{19 x^{11}}{11}+\frac{19 x^9}{9}-\frac{19 x^7}{7}+\frac{19 x^5}{5}-\frac{19 x^3}{3}+19 x \\ \end{array} \right)$$

Claude Leibovici
  • 260,315