query: Integrating : $\int_0^1 {\frac {x^a-x^b} {\ln x} dx}$
positive_passages.docid: 152648
positive_passages.text: Is the integral $\int_1^\infty\frac{x^{-a} - x^{-b}}{\log(x)}\,dx$ convergent?
negative_passages.docid: 132388
negative_passages.text: Calculate $I=\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx$
negative_passages.docid: 63956
negative_passages.text: Evaluate $\int_0^{\infty}\frac{\ln x}{x^a(x+1)}dx$ where $0<a<1$
negative_passages.docid: 765
negative_passages.text: $\int_0^\infty \frac{\log(x)}{x^2+\alpha^2}$ using residues
negative_passages.docid: 41416
negative_passages.text: Integrating $ \int_2^4 \frac{ \sqrt{\ln(9-x)} }{ \sqrt{\ln(9-x)}+\sqrt{\ln(x+3)} } dx. $
negative_passages.docid: 86857
negative_passages.text: Integrate $\int_{0}^{2\pi}\log|1-e^{i\theta}| d\theta$
negative_passages.docid: 115992
negative_passages.text: Integrating $\frac{\log(1+x)}{1+x^2}$
negative_passages.docid: 85949
negative_passages.text: Integral involving logarithm: $\int_0^\infty \frac{ \ln x}{(x+a)(x+b)} dx$
negative_passages.docid: 139784
negative_passages.text: Computing integration $\int_0^1\frac{\ln x}{1-x}d x$
negative_passages.docid: 140638
negative_passages.text: Integrating $\int_0^{\infty} \frac{dx}{1+x^3}$ using residues.
negative_passages.docid: 31964
negative_passages.text: Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$
negative_passages.docid: 79466
negative_passages.text: Evaluating $\int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2}$
negative_passages.docid: 104691
negative_passages.text: Integration of $\int_{0}^{1} \frac{x^k}{1-x} dx$
negative_passages.docid: 99753
negative_passages.text: How to prove $\int_0^1\frac{1-x}{(\ln x)(1+x)}\ dx=\ln\left(\frac2{\pi}\right)$?
negative_passages.docid: 139121
negative_passages.text: Evaluating $\int _0 ^1 x^k(\ln x)^m dx$ for integer $k$ and $m$
negative_passages.docid: 157931
negative_passages.text: Contour integration of $\int \frac{dx} {(1+x^2)^{n+1}}$
negative_passages.docid: 63251
negative_passages.text: Evaluation of $\int^1_0\frac{x^b-x^a}{\ln x} d x=\ln{{\frac{1+b}{1+a}}}$