# Is there an error in Wolfram integrator for determine $\int_{0}^{1}\ln{x}\ln{(-\ln{x})}\cdot{\mathrm dx\over 1+x^2}?$

Consider

$$\int_{0}^{1}\ln{x}\ln{(-\ln{x})}\cdot{\mathrm dx\over 1+x^2}=I=-0.468837...\tag1$$

Wolfram integrator gives a closed form $$I={\gamma_1(3/4)-\gamma_1(1/4)\over 16}+C(\gamma+\ln{4})\ne-0.468837...\tag2$$

But the closed form value doesn't match with the integral value.

Catalan's constant;$C=0.9156...$

$$\gamma_1(3/4)-\gamma_1(1/4)=\pi(\gamma+\ln{4})+2\pi\ln\left({\sqrt{2\pi}\cdot{\Gamma(3/4)\over \Gamma(1/4)}}\right)$$

Is there another mal-function in wolfram integrator software?

How can we evaluate the closed form for $(1)$?

• Perhaps give a link to Wolfram Integrator? Commented Feb 26, 2017 at 17:40
• Here is the link wolframalpha.com/input/… Commented Feb 26, 2017 at 17:43
• Maybe you must ask this question here. I checked and the two expressions are equivalent...
– user173262
Commented Feb 26, 2017 at 17:52
• Wolfram Alpha gives a symbolic expression Commented Feb 26, 2017 at 17:55
• The exponents in $\text{StieltjesGamma}^{(0,1)}(\ldots)$ mean that a derivative is involved. Commented Feb 26, 2017 at 18:25

Let we set $x=e^{-t}$. Then we are looking for a closed form for
$$\int_{0}^{+\infty} e^{-t} (-t \log t)\frac{dt}{1+e^{-2t}}\,dt = \sum_{n\geq 0}(-1)^n \int_{0}^{+\infty}(-t\log t)e^{-(2n+1)t}\,dt$$ and by differentiation under the integral sign $$\int_{0}^{+\infty} (t \log t)e^{-(2n+1)t}\,dt = \left.\frac{d}{d\alpha}\left(\frac{\Gamma(\alpha+1)}{(2n+1)^{\alpha+1}}\right)\right|_{\alpha=1}$$ hence the previous integral equals $$\sum_{n\geq 0}(-1)^n \frac{(\gamma-1)+\log(2n+1)}{(2n+1)^2}=(\gamma-1)K-\frac{1}{4}\left.\frac{d}{d\alpha}\left[\zeta\left(\alpha,\frac{1}{4}\right)-\zeta\left(\alpha,\frac{3}{4}\right)\right]\right|_{\alpha=2}.$$
The integral does not directly depend on the Stieltjes constants $\gamma_1(1/4)$ and $\gamma_1(3/4)$, but rather on $\gamma_1'(\alpha)=\frac{d}{d\alpha}\gamma_1(\alpha)$ at $\alpha=\frac{1}{2}\pm\frac{1}{4}$.