OK, so maybe it's time for a trickier one...

Type 11:\(\displaystyle \mathbb{T}(m\,;\, n)=\int_{0}^{\pi/4}\log^n(\tan x)\log^m[-\log(\tan x)]\,dx\)

Which has the equivalent algebraic form

\(\displaystyle \mathbb{T}(m\,;\, n)=\int_0^1\frac{(\log x)^n\log^m(-\log x)}{1+x^2}\,dx\)

N.B. the following (excellent) article covers similar integrals - although Adamchik seems to have missed the far simpler proof I'll outline below...

http://www.cs.cmu.edu/~adamchik/articles/issac/issac97.pdf Proof:Actually, this one's not anywhere near as scary as it looks... Start with the hyperbolic integral

\(\displaystyle \mathcal{I}(p)=\int_0^{\infty}\frac{x^{p-1}}{\cosh x}\,dx=2\,\int_0^{\infty}\frac{x^{p-1}\,e^{-x}}{(1+e^{-2x})}\,dx=\)

\(\displaystyle 2\,\sum_{k=0}^{\infty}(-1)^k\int_0^{\infty}x^{p-1}\,e^{-(2k+1)x}\,dx\)

for the parameter \(\displaystyle p\in \mathbb{Q}^+\)

Make the substitution \(\displaystyle (2k+1)x=y\,\) to obtain

\(\displaystyle 2\,\sum_{k=0}^{\infty}(-1)^k\int_0^{\infty}x^{p-1}\,e^{-(2k+1)x}\,dx=\)

\(\displaystyle 2\,\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)}\int_0^{\infty}\left(\frac{y}{2k+1}\right)^{p-1}\,e^{-y}\,dy=\)

\(\displaystyle 2\,\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^p}\int_0^{\infty}x^{p-1}\,e^{-x}\,dy=2\,\Gamma(p)\,\beta(p)\)

Where \(\displaystyle \beta(x)\,\) is the Dirichlet Beta Function defined by

\(\displaystyle \beta(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^x}\)

which has the special values

\(\displaystyle \beta(1)=\pi/4\)

\(\displaystyle \beta(2)=G\)

\(\displaystyle \beta(m)=\text{Ti}_m(1)\)

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On the other hand, returning to

\(\displaystyle \mathcal{I}(p)=2\,\int_0^{\infty}\frac{x^{p-1}\,e^{-x}}{(1+e^{-2x})}\,dx=2\,\Gamma(p)\,\beta(p)\)

and setting \(\displaystyle y=e^{-x}\,\) yields

\(\displaystyle \int_0^{\infty}\frac{x^{p-1}\,e^{-x}}{(1+e^{-2x})}\,dx=\Gamma(p)\,\beta(p)=\int_0^1\frac{(-\log y)^{p-1}}{(1+y^2)}\,dy\)

Differentiating both sides m-times w.r.t. the parameter p gives

\(\displaystyle \sum_{j=0}^m\binom{m}{j}\Gamma^{(m-j)}(p)\,\beta^{(j)}(p)=\int_0^1\frac{(-\log y)^{p-1}\log^m(-\log y)}{(1+y^2)}\,dy=\)

\(\displaystyle \int_{0}^{\pi/4}\log^{p-1}(\tan x)\log^m[-\log(\tan x)]\,dx=\mathbb{T}(m\,;\, p-1)\)

And thus we obtain the final result, a binomial sum of derivatives of the Gamma and Dirichlet Beta Functions...

\(\displaystyle \mathbb{T}(m\,;\, n)=\int_{0}^{\pi/4}\log^n(\tan x)\log^m[-\log(\tan x)]\,dx=\)

\(\displaystyle \sum_{j=0}^m\binom{m}{j}\Gamma^{(m-j)}(n+1)\,\beta^{(j)}(n+1)\,.\Box\)