Integrate It:

Calculus Level 4

∫ 0 1 ln ⁡ ( 1 − x 2 ) x d x = ? \large \int_{0}^{1} \frac{\ln(1-x^2)}{x} \, dx = \ ?

Answer upto 3 decimal places


The answer is -0.82246.

Aditya Narayan Sharma by Aditya Narayan Sharma , 21, India

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1 solution

Let I = ∫ 0 1 l n ( 1 − x 2 ) x d x \displaystyle \large \mathfrak{I} = \int_{0}^ {1} \frac{ln(1-x^2)}{x}dx

Using Taylor series expansion of l n ( 1 − x 2 ) ln(1-x^2) ,

I = − ∫ 0 1 ∑ r = 1 ∞ x 2 r x r d x \displaystyle \large\mathfrak{I}=-\int_{0}^{1} \sum_{r=1}^{\infty}\frac{x^{2r}}{xr}dx

I = − ∑ r = 1 ∞ 1 r ∫ 0 1 x 2 r − 1 ⏟ d x \displaystyle \large\mathfrak{I} = -\sum_{r=1}^{\infty}\frac{1}{r} \color{#3D99F6}{\underbrace{\int_{0}^{1} x^{2r-1}}} dx

I = − ∑ r = 1 ∞ 1 r 1 2 r ⏟ \displaystyle \large\mathfrak{I} = -\sum_{r=1}^{\infty} \frac{1}{r} \color{#3D99F6}{\underbrace{\frac{1}{2r}}}

I = − 1 2 ∑ r = 1 ∞ 1 r 2 \displaystyle \large\mathfrak{I} = -\frac{1}{2}\sum_{r=1}^{\infty}\frac{1}{r^2}

I = − ζ ( 2 ) 2 ≈ − 0.82246 \displaystyle \large\mathfrak{I} = -\frac{\zeta(2)}{2}\approx\boxed{-0.82246}

8 Helpful 0 Interesting 1 Brilliant 0 Confused

This is exactly how I did it but I thought ζ ( 2 ) = π 4 \zeta(2) = \frac{\pi}{4} and not π 2 6 \frac{\pi^2}{6}

Josh Banister - 5 years, 1 month ago

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But you got it right , Great !

Aditya Narayan Sharma - 5 years, 1 month ago

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Except I didn't

Josh Banister - 5 years, 1 month ago

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