MIT 2015 Integration Question

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!

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}

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.