So one of the question on the MIT Integration bee has baffled me all day today $$\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x$$ I have tried a variety of things to do this, starting with Integration By Parts Part 1 $$\frac{\tan x-x}{1+x^2}\bigg\rvert_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\frac{-2x(\tan x -x)}{\left (1+x^2 \right )^2}\text{d}x$$ which that second integral is not promising, so then we try Integration By Parts Part 2 $$\tan^{-1} x\tan^2 x\bigg\rvert_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}2\tan^{-1} x\tan x\sec^2 x\text{d}x$$ which also does not seem promising Trig Substitution $x=\tan\theta$ which results $$\int_{0}^{\tan^{-1}\frac{\pi}{4}}\tan^2 \left (\tan\theta\right )\text{d}\theta$$ which I think too simple to do anything with (which may or may not be a valid reason for stopping here) I had some ideas following this like power reducing $\tan^2 x=\frac{1-\cos 2x}{1+\cos 2x}$ which didn't spawn any new ideas. Then I thought maybe something could be done with differentiation under the integral but I could not figure out how to incorporate that. I also considered something with symmetry somehow which availed no results. I'm also fairly certain no indefinite integral exists. Now the answer MIT gave was $\frac{1}{3}$ but wolfram alpha gave $\approx$ .156503. Note The integral I gave was a simplified version of the original here is the original in case someone can do something with it $$\int_{0}^{\frac{\pi}{4}}\frac{1-x^2+x^4-x^6...}{\cos^2 x+\cos^4 x+\cos^6 x...}\text{d}x$$ My simplification is verifiably correct, I'd prefer no complex analysis and this is from this Youtube Video close to the end.

  • $\begingroup$ Your bounds on the integral with trig sub should be $[0,\arctan(\frac{\pi}{4})]$. It is true $\tan (\frac{\pi}{4})=1$ but $\arctan(\frac{\pi}{4}) \neq 1$. $\endgroup$ – Ahmed S. Attaalla Feb 11 '17 at 0:58
  • $\begingroup$ That is true, but at this point it doesn't matter lol @AhmedS.Attaalla $\endgroup$ – Teh Rod Feb 11 '17 at 1:00
  • $\begingroup$ I cannot find the link now, but I'm pretty sure this integral was discussed here on Math.SE last Autumn. I don't remember the outcome of that discussion, though. $\endgroup$ – mickep Feb 15 '17 at 12:13

There must be some problem here. Note that on $[0, \pi/4]$, $$\frac{\tan^2 x}{1+x^2} \le \frac{\tan^2 x}{1 + 0^2} = \tan^2 x,$$ therefore the definite integral is bounded above as follows: $$0 \le \int_{x=0}^{\pi/4} \frac{\tan^2 x}{1+x^2} \, dx < \int_{x=0}^{\pi/4} \tan^2 x \, dx = 1 - \frac{\pi}{4} < \frac{1}{3}.$$ Mathematica is correct; the definite integral cannot be $1/3$. I also watched the YouTube video (see time stamp 1:34:34), and you have transcribed the question correctly and performed the correct algebraic transformations.

Okay so at 1:39:04 they reveal the answer as $1/3$. This is very, very obviously wrong. The very first thought to enter my mind was to check the reasonableness of the result by choosing an appropriate bound.

Interestingly, one contestant at 1:38:42 answers with $\pi/16 \approx 0.19635$ which is remarkably accurate for a last-moment guess, certainly much closer to the mark than the official answer!

  • $\begingroup$ I personally think that the top tangent should've been an arctan which would make this significantly easier $\endgroup$ – Teh Rod Feb 11 '17 at 0:45
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    $\begingroup$ For what it's worth, I attended a different "Institute of Technology," at which, I should imagine, mistakes like this are not so easily made.... $\endgroup$ – heropup Feb 11 '17 at 1:09
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    $\begingroup$ If it exists, Mathematica is not able to find a symbolic expression for its value. And I can count on one hand the number of definite integrals that I can evaluate that it cannot. $\endgroup$ – heropup Feb 11 '17 at 1:16
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    $\begingroup$ Incidentally, $\frac13$ is a factor in the symbolic value of the modified integral where we replaced $\tan$ with $\arctan$: $$\int\limits_0^{\frac\pi4}\frac{\arctan^2 x}{1+x^2}=\frac13\arctan^3\frac\pi4.$$ $\endgroup$ – Ruslan Feb 11 '17 at 8:55
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    $\begingroup$ @JamalS A very simple example is $$\int_{x=0}^{\pi/2} \frac{dx}{1+\tan^{\sqrt{2}} x} = \frac{\pi}{4}.$$ $\endgroup$ – heropup Feb 11 '17 at 15:55

This is not the whole answer to this question but it is something new that wasn't mentioned here so I thought I would post it. Let us denote the unknown integral by $J$. Then substituting for $y= \tan(x)$ we get: \begin{equation} J = \int\limits_0^1 \frac{y^2}{1+\arctan(y)^2} \cdot \frac{1}{1+y^2} d y = \int\limits_0^1 \frac{1}{1+\arctan(y)^2} d y - \int\limits_0^1 \frac{1}{1+\arctan(y)^2} \cdot \frac{dy}{1+y^2}= \left(\sum\limits_{n=0}^\infty (-1)^n \int\limits_0^1 \arctan(y)^{2 n} d y\right) - \arctan(\frac{\pi}{4}) \end{equation} Now, clearly the series converges which follows from the estimate $\arctan(y) < y$ for $y\in(0,1)$. Integrating by parts we have: \begin{eqnarray} \int\limits_0^1 \arctan(y)^2 d y &=& (\frac{\pi}{4})^2+ \frac{\pi}{4} \log(2) - G \\ \int\limits_0^1 \arctan(y)^{2 n} d y &=& \cdots \end{eqnarray} where $G$ is the Catalan constant. Now, of course the real challenge is to calculate the last integral above in closed form. This requires more work. However I have a feeling that this integral can be reduced to poly-logarithms for arbitrary values of $n$ and as such does have a "closed form".

We use the following identities: \begin{eqnarray} [\arctan(y)]^2 &=& \frac{1}{2} \sum\limits_{n=0}^\infty (-1)^n \frac{(y^2)^{n+1}}{n+1} \cdot \left( \Psi(-1/2) + \Psi(n+3/2)\right) \\ &=& -\frac{1}{4} \log ^2\left(\frac{y+i}{y-i}\right)+\frac{1}{2} i \pi \log \left(\frac{y+i}{y-i}\right)+\frac{\pi ^2}{4} \end{eqnarray} Now, raising the identity above to the $n$th power and integrating we readily see that the only non-trivial integral we are dealing is, is the following: \begin{eqnarray} {\mathcal A}_n &:=& \int\limits_0^1 [\log(\frac{x+\imath}{x-\imath})]^n dx \\ &=& (-2 \imath) \int\limits_{-1}^{\imath} \frac{ [\log(u)]^n}{(1-u)^2} d u \\ &=&(-\imath)(\imath \pi)^n + (1+\imath) (\imath \frac{\pi}{2})^n + \\ &&(-\imath)n! (-1)^n \left((\log(2)+\imath \frac{\pi}{2}) 1_{n=1} + 2(S_{1,n-1}(2) - S_{1,n-1}(1-\imath)) 1_{n > 1}\right) \end{eqnarray} where the path in middle integral is a quarter of a unit circle starting at $-1$ and ending at $\imath$. The quantities $S_{1,n-1}()$ in the bottom formula are the Nielsen generalized poly-logarithms. Here $n\ge 1$. Combining the two identities above we obtain the following identity : \begin{eqnarray} &&\int\limits_0^1 \arctan(y)^{2 n} dy = \\ && \left(\frac{\pi^2}{16}\right)^n + \\ && Re\sum\limits_{0\le p_1 \le p_2 \le n} \frac{n! (2(p_2-p_1)+n-p_2)!}{p_1! (p_2-p_1)!(n-p_2)!} \left(\frac{\pi^2}{4}\right)^{p_1} \left(\frac{1}{4}\right)^{p_2-p_1} \left(\frac{\pi}{2}\right)^{n-p_2} \cdot \\ && \left( 0 \cdot 1_{n+p_2-2 p_1=0} + (\log(2)) \cdot 1_{n+p_2-2 p_1=1} + 2 (-\imath)^{n+p_2-2 p_1+1} (S_{1,n+p_2-2 p_1-1}(2) - S_{1,n+p_2-2 p_1-1}(1-\imath)) \cdot 1_{n+p_2-2 p_1>1} \right) \end{eqnarray} valid for $n=1,2,\cdots$. This concludes the calculation.


To avoid the use of complex analysis as done above by @Przemo, could one simply use taylor series expansions and Cauchy products to obtain: \begin{array}{rcl} \displaystyle{\int_0^{\frac{\pi}{4}}\!\!\frac{\tan^2\!x}{1+x^2}dx} & = & \displaystyle{\int_0^{\frac{\pi}{4}}\!\!\tan^2\!x\left(1-x^2+x^4-x^6+\cdots\right)dx} \\[5mm] & = & \displaystyle{\int_0^{\frac{\pi}{4}}\!\!\tan^2\!x\left(\sum_{k=0}^{\infty}(-1)^kx^{2k}\right)dx} \\[5mm] & = & \displaystyle{\int_0^{\frac{\pi}{4}}\!\!\left(\sum_{k=0}^{\infty}\frac{B_{2k+2}(-4)^{k+1}(1-4^{k+1})}{(2k+2)!}x^{2k+1}\right)^{\!\!2}\left(\sum_{k=0}^{\infty}(-1)^kx^{2k}\right)dx} \\[5mm] & = & \displaystyle{\int_0^{\frac{\pi}{4}}\!\!\left(\sum_{k=0}^{\infty}\sum_{l=0}^k\frac{B_{2l+2}B_{2k-2l+2}(-4)^{k+2}(1-4^{l+1})(1-4^{k-l+1})}{(2l+2)!(2k-2l+2)!}x^{2k+2}\right)\left(\sum_{k=0}^{\infty}(-1)^kx^{2k}\right)dx} \\[5mm] & = & \displaystyle{\int_0^{\frac{\pi}{4}}\!\!\sum_{k=0}^{\infty}\sum_{l=0}^k\sum_{m=0}^l\frac{B_{2m+2}B_{2l-2m+2}(-4)^{l+2}(1-4^{m+1})(1-4^{l-m+1})}{(2m+2)!(2l-2m+2)!}(-1)^{k-l}x^{2k+2}dx} \\[5mm] & = & \displaystyle{\sum_{k=0}^{\infty}\sum_{l=0}^k\sum_{m=0}^l\frac{B_{2m+2}B_{2l-2m+2}(-4)^{l+2}(1-4^{m+1})(1-4^{l-m+1})}{(2m+2)!(2l-2m+2)!}(-1)^{k-l}\int_0^{\frac{\pi}{4}}x^{2k+2}dx} \\[5mm] & = & \displaystyle{\sum_{k=0}^{\infty}\sum_{l=0}^k\sum_{m=0}^l\frac{B_{2m+2}B_{2l-2m+2}(-4)^{l+2}(1-4^{m+1})(1-4^{l-m+1})}{(2m+2)!(2l-2m+2)!}(-1)^{k-l}\frac{\frac{\pi}{4}^{2k+3}}{2k+3}} \\[5mm] \end{array}


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