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Sedbober [7]
3 years ago
7

Pls the primitive of this function

Mathematics
1 answer:
ipn [44]3 years ago
8 0

Let <em>F</em> (<em>n</em>) denote the integral,

∫ <em>x</em> (1 - ln(<em>x</em>))<em>ⁿ</em> d<em>x</em>

We attempt to find a power-reduction formula for <em>F</em> (<em>n</em>) in terms of <em>F</em> (<em>n</em> - 1). Integrate by parts, with

<em>u</em> = (1 - ln(<em>x</em>))<em>ⁿ</em>   →   d<em>u</em> = - <em>n</em>/<em>x</em> (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻¹ d<em>x</em>

d<em>v</em> = <em>x</em> d<em>x</em>   →   <em>v</em> = 1/2 <em>x</em> ²

Then

<em>F</em> (<em>n</em>) = <em>u v</em> - ∫ <em>v</em> d<em>u</em>

<em>F</em> (<em>n</em>) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> + <em>n</em>/2 ∫ <em>x</em> (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻¹ d<em>x</em>

<em>F</em> (<em>n</em>) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> + <em>n</em>/2 <em>F</em> (<em>n</em> - 1)

From this relation, we get

<em>F</em> (<em>n</em> - 1) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻¹ + (<em>n</em> - 1)/2 <em>F</em> (<em>n</em> - 2)

<em>F</em> (<em>n</em> - 2) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻² + (<em>n</em> - 2)/2 <em>F</em> (<em>n</em> - 3)

<em>F</em> (<em>n</em> - 3) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻³ + (<em>n</em> - 3)/2 <em>F</em> (<em>n</em> - 4)

and so on, down to

<em>F</em> (1) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>)) + 1/2 <em>F</em> (0)

where

<em>F</em> (0) = ∫ <em>x</em> d<em>x</em> = 1/2 <em>x</em> ² + <em>C</em>

By recursively substituting, we find

→   <em>F</em> (<em>n</em>) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> + <em>n</em>/2 [1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻¹ + (<em>n</em> - 1)/2 <em>F</em> (<em>n</em> - 2)]

… = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> + <em>n</em>/2² <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻¹ + (<em>n</em> (<em>n</em> - 1))/2² <em>F</em> (<em>n</em> - 2)

→   <em>F</em> (<em>n</em>) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> + <em>n</em>/2² <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻¹ + (<em>n</em> (<em>n</em> - 1))/2² [1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻² + (<em>n</em> - 2)/2 <em>F</em> (<em>n</em> - 3)]

… = <em>F</em> (<em>n</em>) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> + <em>n</em>/2² <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻¹ + (<em>n</em> (<em>n</em> - 1))/2³ <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻² + (<em>n</em> (<em>n</em> - 1) (<em>n</em> - 2))/2³ <em>F</em> (<em>n</em> - 3)

→   <em>F</em> (<em>n</em>) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> + <em>n</em>/2² <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻¹ + (<em>n</em> (<em>n</em> - 1))/2³ <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻² + (<em>n</em> (<em>n</em> - 1) (<em>n</em> - 2))/2³ [1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻³ + (<em>n</em> - 3)/2 <em>F</em> (<em>n</em> - 4)]

… = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> + <em>n</em>/2² <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻¹ + (<em>n</em> (<em>n</em> - 1))/2³ <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻² + (<em>n</em> (<em>n</em> - 1) (<em>n</em> - 2))/2⁴ <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻³ + (<em>n</em> (<em>n</em> - 1) (<em>n</em> - 2) (<em>n</em> - 3))/2⁴ <em>F</em> (<em>n</em> - 4)

and so on, down to

<em>F</em> (<em>n</em>) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> + <em>n</em>/2² <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻¹ + … + (<em>n</em> (<em>n</em> - 1) … 2 × 1)/2<em>ⁿ</em> <em>F</em> (0)

<em>F</em> (<em>n</em>) = 1/2 <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> + <em>n</em>/2² <em>x</em> ² (1 - ln(<em>x</em>))<em>ⁿ</em> ⁻¹ + … + (<em>n</em> (<em>n</em> - 1) … 2 × 1)/2<em>ⁿ</em> ⁺¹ <em>x</em> ² + <em>C</em>

We can write this more compactly as the sum,

F(n)=\displaystyle\int f_n(x)\,\mathrm dx=x^2\sum_{k=0}^n \frac{n!}{2^{k+1} (n-k)!} (1-\ln(x))^{n-k} + C

or

F(n)=\displaystyle\int f_n(x)\,\mathrm dx=x^2\sum_{k=0}^n \frac{k!}{2^{k+1}}\binom nk(1-\ln(x))^{n-k} + C

where \binom nk=\frac{n!}{k!(n-k)!} is the binomial coefficient.

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