I tried to solve this indefinite integral, $$\int\frac{1}{1+\tan^{-1}x}\,\text{d}x.$$

I tried taking the change of variable $u=\tan^{-1}x$ but failed to reach a solution.

Can anyone help me?

Thanks in advance.

  • 3
    $\begingroup$ I don't think this can be evaluated in terms of elementary functions (see wolframalpha.com/input/…). $\endgroup$ – Alex Wertheim Mar 13 '14 at 15:56
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    $\begingroup$ And removing the constant term in the denominator doesn't make any difference... You're in NoNameFunctionLand! $\endgroup$ – colormegone Mar 25 '14 at 23:27
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    $\begingroup$ On the other hand, it is amazing to notice that $\int_0^a\frac{dx}{1+\tan^{-1}x}$ is almost linear with $a$ $\endgroup$ – Claude Leibovici Jul 8 '15 at 6:28
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    $\begingroup$ @ClaudeLeibovici It is amazing at first, but then you realise it's trivial due to the asymptotic behaviour of the integrand. $\endgroup$ – Jack Tiger Lam Mar 12 '17 at 3:29
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    $\begingroup$ @ClaudeLeibovici The denominator approaches $1+\frac{\pi}{2}$ as $x$ goes to infinity, so the integral is asymptotically parallel to the line $y=\frac{x}{1+\frac{\pi}{2}}$ $\endgroup$ – Jack Tiger Lam Mar 12 '17 at 3:42

This is not a complete answer, but just converts it to a nested infinite sum.

Letting $f(x) = \frac{1}{1+\tan^{-1}(x)}$, I found the first couple of derivatives at $x = 0$ to be: $$f(0) = 1, f'(0)=-1, f''(0)=2, f'''(0)=-4, f^{(4)}(0)=8, f^{(5)}(0) = -24$$

Plugging these into the OEIS, I found A191700. Using the formula in the OEIS of $$f^{(n)}(0) = (-1)^n n! \sum_{k=1}^{n} k! (-1)^{(3n+k)/2}\sum_{i=k}^{n} \frac{2^{i-k}}{i!} S(i, k) \binom{n-1}{i-1} $$

means that $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = \sum_{n=0}^\infty \left((-1)^n \sum_{k=1}^{n} k! (-1)^{(3n+k)/2}\sum_{i=k}^{n} \frac{2^{i-k}}{i!} S(i, k) \binom{n-1}{i-1} x^n \right)$$

Integrating, we have $$\int f(x) dx = \sum_{n=0}^\infty \left(\frac{(-1)^n}{n+1} \sum_{k=1}^{n} k! (-1)^{(3n+k)/2}\sum_{i=k}^{n} \frac{2^{i-k}}{i!} S(i, k) \binom{n-1}{i-1} x^{n+1} \right) + C$$

This may be able to be simplified by switching the order of summation. However, I got nowhere with that. If anyone makes more progress in simplifying this, feel free to add another answer or post a comment.


Let $u=\tan^{-1}x$ ,

Then $x=\tan u$


$=\int\dfrac{d(\tan u)}{u+1}$

$=\dfrac{\tan u}{u+1}-\int\tan u~d\left(\dfrac{1}{u+1}\right)$

$=\dfrac{\tan u}{u+1}+\int\dfrac{\tan u}{(u+1)^2}~du$

$=\dfrac{\tan u}{u+1}+\int\sum\limits_{n=0}^\infty\dfrac{8u}{((2n+1)^2\pi^2-4u^2)(u+1)^2}~du$ (use Mittag-Leffler Expansion of tangent)

$=\dfrac{\tan u}{u+1}-\int\sum\limits_{n=0}^\infty\dfrac{8}{((2n+1)\pi-2)^2((2n+1)\pi+2u)}~du+\int\sum\limits_{n=0}^\infty\dfrac{8}{((2n+1)\pi+2)^2((2n+1)\pi-2u)}~du+\int\sum\limits_{n=0}^\infty\dfrac{8((2n+1)^2\pi^2+4)}{((2n+1)^2\pi^2-4)^2(u+1)}~du-\int\sum\limits_{n=0}^\infty\dfrac{8}{((2n+1)^2\pi^2-4)(u+1)^2}~du$

$=\dfrac{\tan u}{u+1}-\sum\limits_{n=0}^\infty\dfrac{4\ln((2n+1)\pi+2u)}{((2n+1)\pi-2)^2}-\sum\limits_{n=0}^\infty\dfrac{4\ln((2n+1)\pi-2u)}{((2n+1)\pi+2)^2}+\sum\limits_{n=0}^\infty\dfrac{8((2n+1)^2\pi^2+4)\ln(u+1)}{((2n+1)^2\pi^2-4)^2}+\sum\limits_{n=0}^\infty\dfrac{8}{((2n+1)^2\pi^2-4)(u+1)}+C$


Check by Wolfram Alpha https://www.wolframalpha.com/input/?i=sum+ln%28%282n%2B1%29pi%2Ba%29%2F%28%282n%2B1%29pi-2%29%5E2%2Cn%3D0+to+inf and https://www.wolframalpha.com/input/?i=sum+ln%28%282n%2B1%29pi-a%29%2F%28%282n%2B1%29pi%2B2%29%5E2%2Cn%3D0+to+inf, the two infinite series converges.

  • $\begingroup$ I don't think that will work as in your second sum, $2u-(2n+1)\pi$ will have a negative argument inside the $\ln$. $\endgroup$ – automaticallyGenerated Sep 27 '19 at 5:43

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