Let's do some calculus! (50)

Calculus Level 5

∫ 0 1 { ( − 1 ) ⌊ 1 / x ⌋ ⋅ 1 x } d x \large \int_0^1 \left\{(-1)^{\left\lfloor 1/x \right\rfloor}\cdot \frac 1x \right\} \ \mathrm{d}x

If the value of the above integral can be represented as a + ln ⁡ ( b π ) a + \ln \left( \dfrac b\pi \right) for positive integers a a and b b , find a + b a+b .

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The answer is 3.

Tapas Mazumdar by Tapas Mazumdar , 20, India

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1 solution

First Last
Jun 15, 2017

∫ 0 1 { ( − 1 ) ⌊ 1 x ⌋ ⋅ 1 x } d x = ∫ 1 ∞ ( − 1 ) ⌊ x ⌋ x − ⌊ ( − 1 ) ⌊ x ⌋ x ⌋ x 2 d x by 1 x → x \displaystyle \int_0^1 \left\{(-1)^{\left\lfloor \frac{1}{x}\right\rfloor}\cdot\dfrac{1}{x}\right\}dx = \int_{1}^{\infty}\frac{(-1)^{\lfloor x\rfloor}x - \lfloor (-1)^{\lfloor x\rfloor}x\rfloor}{x^2}dx\quad\text{ by }\frac{1}{x}\rightarrow x

Split into regions of floor(odd.xyz) = odd and floor(even.xyz) = even

∑ n = 1 ∞ ∫ 2 n − 1 2 n − 1 x d x + ∫ 2 n 2 n + 1 1 x d x − ∑ n = 1 ∞ ∫ 2 n − 1 2 n − 2 n x 2 d x + ∫ 2 n 2 n + 1 2 n x 2 d x = \displaystyle\sum_{n=1}^{\infty}\int_{2n-1}^{2n}\frac{-1}{x}dx+\int_{2n}^{2n+1}\frac{1}{x}dx-\sum_{n=1}^{\infty}\int_{2n-1}^{2n}\frac{-2n}{x^2}dx+\int_{2n}^{2n+1}\frac{2n}{x^2}dx =

∑ n = 1 ∞ − ln ⁡ ( 2 n ) + ln ⁡ ( 2 n − 1 ) + ln ⁡ ( 2 n + 1 ) − ln ⁡ ( 2 n ) + ∑ n = 1 ∞ 2 n 2 n − 1 + 2 n 2 n + 1 − 2 = \displaystyle\sum_{n=1}^{\infty}-\ln{(2n)}+\ln{(2n-1)}+\ln{(2n+1)}-\ln{(2n)} + \sum_{n=1}^{\infty}\frac{2n}{2n-1}+\frac{2n}{2n+1}-2 =

ln ⁡ ( ∏ n = 1 ∞ 1 − 1 4 n 2 ) + ∑ n = 1 ∞ 1 2 n − 1 − 1 2 n + 1 \ln{\big( \prod_{n=1}^{\infty}1-\frac1{4n^2}\big)} + \sum_{n=1}^{\infty}\frac{1}{2n-1}-\frac{1}{2n+1}

∏ n = 1 ∞ 1 − 1 4 n 2 = 2 π \displaystyle\prod_{n=1}^{\infty}1-\frac1{4n^2} = \boxed{\frac{2}{\pi}} from the Wallis Product

∑ n = 1 ∞ 1 2 n − 1 − 1 2 n + 1 = ∑ n = 1 ∞ ∫ 0 1 ( y 2 ) n − 1 d y − ( y 2 ) n d y = ∫ 0 1 ∑ n = 1 ∞ ( y 2 ) n − 1 d y − ( y 2 ) n d y = \displaystyle\sum_{n=1}^{\infty}\frac{1}{2n-1}-\frac{1}{2n+1} = \sum_{n=1}^{\infty}\int_{0}^{1}(y^2)^{n-1}dy-(y^2)^ndy =\int_{0}^{1}\sum_{n=1}^{\infty}(y^2)^{n-1}dy-(y^2)^ndy =

∫ 0 1 1 1 − y 2 − ( 1 1 − y 2 − 1 ) = 1 \displaystyle\int_{0}^{1}\frac{1}{1-y^2}-( \frac{1}{1-y^2}-1) = \boxed{1}\quad by noticing the infinite geometric sequence and reversing the sum and integral. ln ⁡ 2 π + 1 \displaystyle\ln{\frac{2}{\pi}}+1

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You made a mistake here

∑ n = 1 ∞ ( ∫ 2 n − 1 2 n − 1 x d x + ∫ 2 n 2 n + 1 1 x d x ) = ∑ n = 1 ∞ ( − ln ⁡ ( 2 n ) + ln ⁡ ( 2 n − 1 ) + ln ⁡ ( 2 n + 1 ) − ln ⁡ ( 2 n ) ) = ∑ n = 1 ∞ ln ⁡ ( 4 n 2 − 1 4 n 2 ) = ln ⁡ ( ∏ n = 1 ∞ 4 n 2 − 1 4 n 2 ) = ln ⁡ ( 2 π ) \begin{aligned} \sum_{n=1}^{\infty} \left( \int_{2n-1}^{2n} \dfrac{-1}{x} \, dx + \int_{2n}^{2n+1} \dfrac{1}{x} \,dx \right) &= \sum_{n=1}^{\infty} \left( - \ln (2n) + \ln (2n-1) + \ln(2n+1) - \ln (2n) \right) \\ &= \sum_{n=1}^{\infty} \ln \left( \dfrac{4n^2 -1}{4n^2} \right) \\ &= \ln \left( \prod_{n=1}^{\infty} \dfrac{4n^2 -1}{4n^2} \right) \\ &= \ln \left( \dfrac{2}{\pi} \right) \end{aligned}

which is the reciprocal of the Wallis Product.

Tapas Mazumdar - 3 years, 12 months ago

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Thanks I typed that wrong.

First Last - 3 years, 12 months ago

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