Let's do some calculus! (58)

Calculus Level 4

∫ 0 1 1 x x 1 x ⋯ x ( − 1 ) n ⋯ n + 1 n 4 3 3 2 d x = 1 ln ⁡ a \large \int_0^1 \sqrt{ \dfrac{1}{x} \sqrt[\frac32]{ x \sqrt[\frac43]{ \dfrac{1}{x} \cdots \sqrt[\frac{n+1}{n}]{x^{(-1)^n} \cdots}}}} \, \ dx = \dfrac{1}{\ln a}

where n n runs from 1 1 to ∞ \infty . Find the value of a a .

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The answer is 2.000.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

Guilherme Niedu
May 19, 2018

I = ∫ 0 1 ( x − 1 2 ⋅ x 2 3 ⋅ 1 2 ⋅ x − 3 4 ⋅ 2 3 ⋅ 1 2 ⋅ . . . ) d x \large \displaystyle I = \int_0^1 \left ( x^{-\frac12} \cdot x^{ \frac23 \cdot \frac12} \cdot x^{- \frac34 \cdot \frac23 \cdot \frac12} \cdot ... \right ) dx

I = ∫ 0 1 x ( − 1 2 + 1 3 − 1 4 + . . . ) d x \large \displaystyle I = \int_0^1 x^{\left ( -\frac12 + \frac13 - \frac14 + ... \right ) } dx

Recalling that ln ⁡ ( 2 ) = 1 − 1 2 + 1 3 − 1 4 + . . . \ln(2) = 1 -\frac12 + \frac13 - \frac14 +... :

I = ∫ 0 1 x [ ln ⁡ ( 2 ) − 1 ] d x \large \displaystyle I = \int_0^1 x^{\left [ \ln(2) - 1 \right ] } dx

I = 1 ln ⁡ ( 2 ) \color{#20A900} \boxed { \large \displaystyle I = \frac{1}{\ln(2)} }

Thus:

a = 2 \color{#3D99F6} \boxed { \large \displaystyle a = 2 }

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