Evaluate $$\int_{0}^{1} \ln (x) \ln(1-x) dx$$
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3Two line questions are frowned upon... – Simply Beautiful Art May 20 '17 at 14:08
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the result should be $$2-\frac{\pi^2}{6}$$ – Dr. Sonnhard Graubner May 20 '17 at 14:10
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5Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove," "Evaluate," etc.) to be rude when asking for help; please consider rewriting your post. – Clement C. May 20 '17 at 14:11
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4If I may suggest, just ignore any answer or comment which gives the result without any explanation. Anyone is able to use a CAS. I suppose that you want to learn. – Claude Leibovici May 20 '17 at 14:24
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Learning to use a CAS is even better. – Brethlosze May 20 '17 at 14:31
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Possible duplicate of Evaluate $ \int_{0}^{1} \ln(x)\ln(1-x),dx $ – Zaid Alyafeai May 20 '17 at 21:41
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$$\int_0^1 \ln(x) \ln(1-x) dx = -\int_0^1 \ln(x) \sum_{n=1}^\infty \frac{x^n}{n} dx \\ =-\sum_{n=1}^\infty \frac{1}{n} \int_0^1 x^n \ln(x) dx $$ So, let $I_n = \int_0^1 x^n \ln(x) dx$. Using integration by parts, we see $I_n = -\frac{1}{(n+1)^2}$, so $$\int_0^1 \ln(x) \ln(1-x) dx = \sum_{n=1}^\infty \frac{1}{n(n+1)^2} \\ = \sum_{n=2}^\infty \frac{1}{(n-1)n^2} \\ = \sum_{n=2}^\infty \frac{1}{n-1} - \frac{1}{n}-\frac{1}{n^2} \\ = \sum_{n=2}^\infty \left(\frac{1}{n-1} - \frac{1}{n}\right)- \sum_{n=2}^\infty\frac{1}{n^2} \\ = \frac{1}{2-1} - \left(\frac{\pi^2}{6} - 1\right) \\ = 2 - \frac{\pi^2}{6} \\$$
B. Mehta
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2You probably should justify why you are allowed to swap $\sum_1^\infty$ and $\int_0^1$ in your second step. This is not always possible. – Clement C. May 20 '17 at 14:30
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