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Calculus Level 4

E = ∫ 0 1 ln ⁡ ( c ) ln ⁡ ( 1 − c ) d c \mathscr{E} = \displaystyle \int_{0}^{1} \ln(c) \ln(1-c) \, dc

If E \mathscr{E} can be represented in the form a − π b d a - \dfrac{\pi^{b}}{d} , where a , b , d a, b, d are positive integers, find a + b + d a + b + d .


The answer is 10.

Hobart Pao by Hobart Pao , 23, USA

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1 solution

This problem is in integration tricks

∫ 0 1 ln ⁡ x ln ⁡ ( 1 − x ) d x = − ∫ 0 1 ∑ k = 1 ∞ x k ln ⁡ x k d x . \int_0^1 \ln x \ln(1-x) \, dx = - \int_0^1 \sum_{k = 1}^\infty \frac{x^k \ln x}{k} \, dx. Since Fubini's Theorem applies here or using Abel theorem and Weierstras theorem for uniform convergenging of continuous functions, this is equal to − ∑ k = 1 ∞ ∫ 0 1 x k ln ⁡ x k d x = ∑ k = 1 ∞ 1 k ( k + 1 ) 2 = ∑ k = 1 ∞ 1 k ( k + 1 ) − ∑ k = 1 ∞ 1 ( k + 1 ) 2 = - \sum_{k = 1}^\infty \int_0^1 \frac{x^k \ln x}{k} \, dx = \sum_{k = 1}^\infty \frac{1}{k (k+1)^2} = \sum_{k = 1}^\infty \frac{1}{k(k + 1)} - \sum_{k = 1}^\infty \frac{1}{(k + 1)^2} = = ∑ k = 1 ∞ ( 1 k − 1 k + 1 ) − ∑ k = 1 ∞ 1 ( k + 1 ) 2 = 1 − ( π 2 6 − 1 ) = 2 − π 2 6 . =\sum_{k = 1}^\infty (\frac{1}{k} - \frac{1}{k + 1}) - \sum_{k = 1}^\infty \frac{1}{(k + 1)^2} = 1 - (\frac{\pi^2}{6} - 1) = \boxed{2 - \frac{\pi^2}{6}.}

Details.-

McLaurin series of ln ⁡ ( 1 − x ) = − ∑ k = 1 ∞ x k k if |x| < 1 \ln(1 - x) = - \sum_{k = 1}^{\infty} \frac{x^k}{k} \quad \quad \text{ if |x| < 1}

Let ( k ≠ − 1 \small k \neq -1 ) I ( k ) = ∫ 0 1 x k d x = 1 k + 1 ⇒ ∂ ∂ k I ( k ) = − 1 ( k + 1 ) 2 = ∫ 0 1 x k ln ⁡ x d x I(k) = \int_0^1 x^k dx = \frac{1}{k + 1} \Rightarrow \frac{\partial}{\partial k} I(k) = \frac{- 1}{(k +1)^2} = \int_0^1 x^k\ln x \space dx

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An easier way would be to differentiate beta function.

A Former Brilliant Member - 3 years, 9 months ago

Can you explain the last line

Kushal Bose - 4 years, 11 months ago

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yes, of course, a little more explained: I ( k ) = ∫ 0 1 x k d x = I(k) = \int_0^1 x^k dx = integrating respect to x = x k + 1 k + 1 ∣ 0 1 = 1 k + 1 − 0 k + 1 = 1 k + 1 , ( 1 ) ⇒ = \frac{x^{k +1}}{k + 1} \bigg|_0^1 = \frac{1}{k + 1} - \frac{0}{k + 1} = \frac{1}{k + 1},\space (1) \Rightarrow Derivating now (1) respect to k ∂ ∂ k I ( k ) = − 1 ( k + 1 ) 2 = ∫ 0 1 x k ln ⁡ x d x = ∫ 0 1 ∂ ∂ k x k d x \frac{\partial}{\partial k} I(k) = \frac{- 1}{(k +1)^2} = \int_0^1 x^k\ln x \space dx = \int_0 ^1 \frac{\partial}{\partial k} x^k \space dx

differentiating under sign of integral

I hope this help you

Guillermo Templado - 4 years, 11 months ago

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