7
$\begingroup$

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.

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

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

Then $x=\tan u$

$\therefore\int\dfrac{1}{1+\tan^{-1}x}~dx$

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

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

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

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

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

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

$=\sec^21\ln(\tan^{-1}x+1)+\dfrac{x+\tan1}{\tan^{-1}x+1}-\sum\limits_{n=0}^\infty\dfrac{4\ln((2n+1)\pi+2\tan^{-1}x)}{((2n+1)\pi-2)^2}-\sum\limits_{n=0}^\infty\dfrac{4\ln((2n+1)\pi-2\tan^{-1}x)}{((2n+1)\pi+2)^2}+C$

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.

$\endgroup$
2
  • $\begingroup$ I don't think that will work as in your second sum, $2u-(2n+1)\pi$ will have a negative argument inside the $\ln$. $\endgroup$ Sep 27 '19 at 5:43
  • $\begingroup$ Try to do ln|negative| which usually helps. $\endgroup$ Aug 5 at 17:19
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ What is S(i,k)? Maybe the Stieltjes constants or maybe they are in the link. $\endgroup$ Aug 5 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.