4
$\begingroup$

I've been playing with the following definite integral and was wondering if anyone knew of any methods to solve?

$$I = \int_{0}^{1} \frac{1}{1 + \arctan(x)} \:dx$$

$\endgroup$
  • $\begingroup$ There doesn't seem to be an elementary antiderivative. $\endgroup$ – Robert Israel Oct 26 '18 at 3:20
  • $\begingroup$ @RobertIsrael Indeed. I do wonder whether though it can be expressed in terms of common non-elementary functions. $\endgroup$ – user150203 Oct 26 '18 at 3:20
1
$\begingroup$

Starting from $$ \cos(x) = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2 \pi^2}\right) \tag{1}$$ and applying $\frac{d^2}{dx^2}\log(\cdot)$ to both sides we get $$ \frac{1}{\cos^2 x} = 8\sum_{n\geq 0}\frac{\pi^2(2n+1)^2+4x^2}{(\pi^2(2n+1)^2-4x^2)^2}\tag{2}$$ By applying $\int_{0}^{\pi/4}\left(\cdot\right)\frac{dx}{x+1}$ we get that $I$ can be represented as a (horrible) series whose general term is $O\left(\frac{\log n}{n^2}\right)$. A numerical approximation of the given integral is better performed through some version of the Shafer-Fink inequality, but the series representation gives that there is a (feeble) relation between the given integral and $\zeta'(2)$, which on its turn is related to the Glaisher-Kinkelin constant. Shafer-Fink in its original formulation leads to

$$ \int_{0}^{1}\frac{dx}{1+\arctan x}\approx \int_{0}^{1}\frac{1+2\sqrt{1+x^2}}{1+3x+2\sqrt{1+x^2}}\,dx=\int_{1}^{1+\sqrt{2}}\frac{(1+t^2)(1+t+t^2)}{t^2(5t^2+2t-1)}\,dt $$ where the RHS can be computed by partial fraction decomposition. It equals $\approx \color{green}{0.714}\color{red}{504} $.

$\endgroup$
  • $\begingroup$ Thank you very much for your post Jack. If I may ask, where does the $\cos(x)$ property come from. $\endgroup$ – user150203 Oct 27 '18 at 14:19
  • 1
    $\begingroup$ @DavidG: it is the Weierstrass product for the cosine function. Any entire with finite order can be factored in a similar way. It is a consequence (on the very long run) of the Mittag-Leffler theorem. $\endgroup$ – Jack D'Aurizio Oct 27 '18 at 14:22
  • $\begingroup$ Thanks for that. I will look into it. Do you think it's possible for the integral to be expressed in terms of known constants and/or elementary and/or non-elementary functions? $\endgroup$ – user150203 Oct 27 '18 at 14:29
  • $\begingroup$ @DavidG: I would not bet on it. Some expert in differential Galois theory might provide a definitive (negative) answer. On the other hand all the integrals of the form $\int_{0}^{1}(\arctan x)^m\,dx$ are related to Euler sums (with increasing weight) / polylogarithms evaluated at roots of unity. $\endgroup$ – Jack D'Aurizio Oct 27 '18 at 14:31
  • $\begingroup$ Hmm, I hope one day to have the skills to know where and when integrals of this nature can be expressed in terms of those features discussed. If I may ask one more question, what would you recommend as a texbook/course/etc to develop those skills? $\endgroup$ – user150203 Oct 27 '18 at 14:34
2
$\begingroup$

The approach I took:

First let $u = \tan(x)$ to yield:

$$I = \int_{0}^{\frac{\pi}{4}} \frac{\tan^2(u) + 1}{1 + u} \:du$$

Unfortunately I had no luck with my usual tactics, so I decided to use the Taylor series of $\tan^2(u)$ at $u = 0$ which I was greatly helped with here.

We find that:

$$ \tan^{2}(u) + 1 = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} u^{2n-2} $$

Which holds for $|u| < \frac{\pi}{2}$. As the domain of the integral is within that, we can use this expansion (I believe).

Hence, we arrive at,

$$I = \int_{0}^{\frac{\pi}{4}} \frac{\sum_{n = 1}^{\infty} C_n u^{2n - 2}}{1 + u} \:du = \sum_{n = 1}^{\infty}C_n\int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{1 + u} \:du = \sum_{n = 1}^{\infty}C_n\: F_n $$

Where

$$ C_n = \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} ,\qquad F_n = \int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{u + 1} \: du$$

Taking the Taylor expansion for $\frac{1}{1 + u}$ at $u = 0$ (as given here) we can evaluate $F_n$:

\begin{align} F_n &= \int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{u + 1} \: du \\ &= \left[\ln|u + 1| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k u^k \right]_{0}^{\frac{\pi}{4}} \\ &= \ln\left|\frac{\pi}{4} + 1 \right| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k \frac{\pi^k}{4^kk} \end{align}

From which we arrive at:

\begin{align} I = \sum_{n = 1}^{\infty} C_n\:F_n &= \sum_{n = 1}^{\infty} C_n \left[\ln\left|\frac{\pi}{4} + 1 \right| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k \frac{\pi^k}{4^kk} \right] \\ &= \ln\left|\frac{\pi}{4} + 1 \right| \sum_{n = 1}^{\infty} C_n + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n - 2} C_n \left(-1\right)^k \frac{\pi^k}{4^kk} \end{align}

Recall that

$$ \tan^{2}(u) + 1 = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} u^{2n-2} = \sum_{n = 1}^{\infty} C_n u^{2n - 2} $$

Hence,

$$ \tan^{2}(1) + 1 = \sec^{2}(1) = \sum_{n = 1}^{\infty} C_n$$

Thus,

\begin{align} I &= \ln\left|\frac{\pi}{4} + 1 \right| \sum_{n = 1}^{\infty} C_n + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n - 2} C_n \left(-1\right)^k \frac{\pi^k}{4^kk} \\ &= \ln\left|\frac{\pi}{4} + 1 \right|\sec^2(1) + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n -2}\frac{\left(-1\right)^kB_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)!4^kk} \end{align}

However at this stage, I'm lost as to how this could be simplified.

$\endgroup$
  • $\begingroup$ I added an edit to my answer. Might help a little $\endgroup$ – clathratus Oct 26 '18 at 4:13
  • 1
    $\begingroup$ @clathratus - Thank you, appreciate the help. $\endgroup$ – user150203 Oct 26 '18 at 4:43
1
$\begingroup$

Let's take a look at that big 'ole fraction. $$Q_{nk}=\frac{\left(-1\right)^kB_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)!4^kk}$$ $$Q_{nk}=\frac{\left(-1\right)^kB_{2n} (-1)^n 4^n\left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)!4^kk}$$ $$Q_{nk}=\frac{\left(-1\right)^{n+k}B_{2n} 4^{n-k}\left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)!k}$$ $$Q_{nk}=\frac{\left(-1\right)^{n+k}B_{2n} 4^{n-k}\left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)(2n-1)(2n-2)!k}$$ $$Q_{nk}=\frac{\left(-1\right)^{n+k}B_{2n} 4^{n-k}\left(1-4^n\right)\pi^k}{(2nk)(2n-2)!}$$ Which is the tiniest bit more compact, but still probably isn't enough.

Edit: here's a little more $$Q_{nk}=\frac{\left(-1\right)^{n+k}B_{2n} 2^{2(n-k)}\left(1-4^n\right)\pi^k}{(2nk)(2n-2)!}$$ $$Q_{nk}=\frac{\left(-1\right)^{n+k}B_{2n} 2^{2n-2k-1}\left(1-4^n\right)\pi^k}{nk(2n-2)!}$$

$\endgroup$

Your Answer

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