Closed form for sum of divergent integrals

I am dealing with this sum of two divergent integrals

$$-\frac{1}{2}\int _0^{1}\frac{\left(\frac{1-t}{1+t}\right)\arctan \left(t\right)}{t\left(\ln t\right)^2}dt-\frac{\pi }{4}\int _0^{1}\frac{1}{\ln \left(t\right)\left(1+t\right)^2}dt$$

This is ugly but although they diverge seperately, their sum converges to $0.1746809...$

Can we find a closed form in terms of known constants for this number?

I've tried getting like denominators by setting $t\to e^{-\sqrt{|\ln(t)|}}$ in the second integral.

Which gave $$-\frac{\pi }{4}\int _0^{1}\frac{\frac{2\left(\ln t\right)e^{\left(\ln t\right)^2}}{\left(1+e^{\left(\ln t\right)^2}\right)^2}}{t\left(\ln t\right)^2}dt$$

However I realized that the expression has to be written in terms of the limit $m\to 1$, because although the upper bound in this integral is $1$, both integrals separately diverge at different rates as their upper bounds go to $1$. So we'd have to write the expression as

$$\lim_{m\to 1}\left(-\frac{1}{2}\int _0^{m}\frac{\left(\frac{1-t}{1+t}\right)\arctan t}{t\left(\ln t\right)^2}dt-\frac{\pi }{4}\int _0^{e^{-\sqrt{|\ln m|}}}\frac{\frac{2\left(\ln t\right)e^{\left(\ln t\right)^2}}{\left(1+e^{\left(\ln t\right)^2}\right)^2}}{t\left(\ln t\right)^2}dt\right)$$

Thus we can't combine these integrals and this doesn't work.

Due to Jack's comment I will rewrite the integral as $$\int _0^{1}\frac{dt}{\ln t}\left(\frac{\pi }{2\left(1+t\right)^2}+\frac{1-t}{1+t}\frac{\arctan t}{t\ln t}\right)$$ which is just twice the original value.

• That has got to be one of the ugliest simple substitutions of the year. – Simply Beautiful Art Aug 9 '17 at 15:22
• Oh really? Then I guess my separate integrals are not each divergent? – tyobrien Aug 9 '17 at 15:54
• They are separately divergent because each integrand is has a $\pm1/(t-1)$ singularity at $1.$ – zhw. Aug 9 '17 at 16:15

We may consider that

$$\int_{0}^{1}\left(\frac{1}{\log t}\color{red}{+\frac{1}{1-t}}\right)\frac{dt}{(1+t)^2}\stackrel{t\mapsto e^{-x}}{=}\int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)\frac{dx}{(e^{-x}+1)^2}$$ where the red term has been introduced to counter-act the singularity at $t=1$.
An integral similar to an integral representation of the Euler-Mascheroni constant appears.

By the (inverse) Laplace transform, the last integral equals

$$\int_{0}^{+\infty}\left(\log(s+1)-\psi(s+1)\right)\cdot\sum_{n\geq 1}(-1)^{n+1}n\,\delta(s-n+1)\,ds$$ or $$\sum_{n\geq 1}n(-1)^{n+1}\left(\log(n)-H_{n-1}+\gamma\right)=\frac{1}{2}+\frac{\gamma}{4}+\frac{7\log 2}{12}-3\log A$$ where the Glaisher-Kinkelin constant $A$ is related to $\zeta'(-1)$ and $\zeta'(2)$.

It just remains to apply a similar approach to the other integral, i.e. to get rid of a pole and to compute the regularized integral in a explicit way, in terms of $\gamma,\pi,\log 2$ and values of $\zeta$ and $\zeta'$.

$$\small \int_{0}^{1}\left(\frac{1-t}{1+t}\cdot\frac{\arctan t}{t\log^2 t}\color{red}{-\frac{\pi}{8(1-t)}}\right)\,dt \stackrel{t\mapsto e^{-x}}{=} \int_{0}^{+\infty}\left(\frac{\pi}{8-8e^x}+\frac{\arctan(e^{-x})(e^x-1)}{x^2 (e^x+1)}\right)\,dx$$ equals $$-\frac{\pi \gamma}{8}+\int_{0}^{+\infty}\left(\frac{\arctan(e^{-x})(e^x-1)}{x^2 (e^x+1)}-\frac{\pi}{8xe^x}\right)\,dx$$ which can be converted in a series by differentiation under the integral sign and Frullani's Theorem. We may clearly see the Gudermannian function being involved in this conversion step:

$$-\frac{\pi \gamma}{8}+\int_{0}^{+\infty}\left(\frac{\pi(e^x-1)}{4x^2(e^x+1)}-\frac{\pi}{8xe^x}\right)\,dx - \int_{0}^{1}\int_{0}^{+\infty}\frac{e^{-kx}(e^x-1)}{x(e^x+1)(e^{-2kx}+1)}\,dx\,dk$$ equals (by the Laplace transform): $$-\frac{\pi\gamma}{8}+\frac{\pi}{6}\log\frac{A^9}{2}- \int_{0}^{1}\int_{0}^{+\infty}\frac{e^{-kx}(e^x-1)}{x (e^x+1)(e^{-2kx}+1)}\,dx\,dk$$ hence the whole problem boils down to computing

$$\int_{0}^{+\infty}\frac{\tanh\left(\frac{x}{2}\right)\arctan \tanh\left(\frac{x}{2}\right)}{x^2}\,dx\qquad \text{or}\qquad \int_{0}^{+\infty}\tanh(x)\arctan\tanh(x)\frac{dx}{x^2}.$$ By the residue theorem we have $$\int_{0}^{+\infty}\frac{\tanh^2 x}{x^2}\,dx = \frac{14\,\zeta(3)}{\pi^2},\qquad \int_{0}^{+\infty}\frac{\tanh^4 x}{x^2}\,dx=\frac{4 \left(14 \pi^2 \,\zeta(3)-93\,\zeta(5)\right)}{3 \pi^4}$$ and the previous integral can be written in terms of the values of the $\zeta$ function.
A numerical evaluation is actually way easier, since $\frac{\tanh(x)\arctan\tanh(x)}{x^2}\approx\frac{1}{x^2+1}$ on $\mathbb{R}^+$.

• Great work! But how can the two integrals with the red terms be combined to equal my value? Their factors are wrong. – tyobrien Aug 10 '17 at 14:53
• @user991003: due to the terms $\frac{1}{2}$ and $\frac{\pi}{4}$ in your original formula, the red terms I introduced to get rid of the singularities cancel out. – Jack D'Aurizio Aug 10 '17 at 14:56
• But your second integral is missing a $1/t$ in it and they can't cancel if the red term in the first integral is multiplied by $\frac{1}{(1+t)^2}$ – tyobrien Aug 10 '17 at 14:58
• @user991003: $$\color{blue}{\frac{\pi}{4}}\int_{0}^{1}\left(\frac{1}{\log t}\color{red}{+\frac{1}{1-t}}\right)\frac{dt}{(1+t)^2}+\color{blue}{\frac{1}{2}}\int_{0}^{1}\left(\frac{1-t}{1+t}\cdot\frac{\arctan t}{t\log^2 t}\color{red}{-\frac{\pi}{8(1-t)}}\right)\,dt$$ gives the opposite of the integral you are interested into, plus an elementary integral. Is it clear now? The above combination equals $$\frac{\pi}{16}(1+\log 2)-\color{orange}{\text{what we care about}}.$$ – Jack D'Aurizio Aug 10 '17 at 15:04
• Oh yeah, I see what you're doing now. Thanks. – tyobrien Aug 10 '17 at 15:20