Integrable?

Calculus Level 4

I = 0 1 ln ( x ) x 1 d x \displaystyle I = \int_{0}^{1}\dfrac{\ln(x)}{x-1} \, dx

Find 100 I 100I upto 3 decimal places.


The answer is 164.493.

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4 solutions

Ravi Dwivedi
Dec 29, 2015

Maclurin expansion of l n ( 1 + x ) = x 1 2 x 2 + 1 3 x 3 1 4 x 4 + . . . ln(1+x) = x - \frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4+... where 1 < x < 1 -1<x<1

Replace x x by x 1 x-1 in above formula

l n ( x ) = ( x 1 ) 1 2 ( x 1 ) 2 + 1 3 ( x 1 ) 3 1 4 ( x 1 ) 4 + . . . ln(x)=(x-1)- \frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+...

Also by limits of integration 0 < x < 1 0<x<1 means 1 < x 1 < 0 -1<x-1<0 So we can use this above formula

Now ln x x 1 = 1 1 2 ( x 1 ) + 1 3 ( x 1 ) 2 1 4 ( x 1 ) 3 + . . . \frac{\ln x}{x-1}=1- \frac{1}{2}(x-1)+\frac{1}{3}(x-1)^2-\frac{1}{4}(x-1)^3+...

I = 0 1 ln ( x ) x 1 d x = 0 1 ( 1 1 2 ( x 1 ) + 1 3 ( x 1 ) 2 1 4 ( x 1 ) 3 + . . . ) d x \displaystyle I = \int_{0}^{1}\dfrac{\ln(x)}{x-1} \, dx = \int_{0}^{1} (1- \frac{1}{2}(x-1)+\frac{1}{3}(x-1)^2-\frac{1}{4}(x-1)^3+... ) \, dx

Using a b f ( x ) d x = a b f ( a + b x ) d x \int_a^b \mathrm f(x)\,\mathrm{d}x = \int_a^b \mathrm f(a+b-x)\,\mathrm{d}x we get

I = 0 1 ( 1 1 2 ( 1 x 1 ) + 1 3 ( 1 x 1 ) 2 1 4 ( 1 x 1 ) 3 + . . . ) d x \displaystyle I = \int_{0}^{1} (1- \frac{1}{2}(1-x-1)+\frac{1}{3}(1-x-1)^2-\frac{1}{4}(1-x-1)^3+... ) \, dx

I = 0 1 ( 1 + 1 2 x + 1 3 x 2 + 1 4 x 3 + . . . ) d x \displaystyle I=\int_{0}^{1} (1+ \frac{1}{2}x+\frac{1}{3}x^2+\frac{1}{4}x^3+... ) \, dx

I = 1 + 1 2 2 + 1 3 2 + 1 4 2 + . . . . = π 2 6 \displaystyle I= 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....=\frac{\pi^{2}}{6}

You can check this above sum here

100 I = 100 π 2 6 = 164.493 100I = 100\frac{\pi^{2}}{6}=\boxed{164.493}

Moderator note:

Nice usage of the symmetric integration trick to simplify the working.

I = 0 1 l n ( x ) 1 x d x \displaystyle I = - \int_{0}^{1} \dfrac{ln(x)}{1-x}dx
Using the Maclaurin series for 1 1 x \dfrac{1}{1-x}
I = 0 1 l n ( x ) × ( 1 + x ² + x ³ . . . ) d x = 0 1 l n ( x ) k = 0 x k d x \displaystyle \therefore I = -\int_{0}^{1} ln(x) \times \left( 1 + x² + x³ ... \right) dx = -\int_{0}^{1} ln(x)\sum_{k= 0}^{\infty}x^{k} dx
Interchanging the order of integration and summation,
I = k = 0 0 1 x k l n ( x ) d x \displaystyle \therefore I = -\sum_{k=0}^{\infty} \int_{0}^{1} x^{k}ln(x)dx
Now, it can be proved that,
0 1 x m l n n ( x ) d x = ( 1 ) n × n ! ( m + 1 ) n + 1 \displaystyle \int_{0}^{1} x^{m}ln^{n}(x)dx = \dfrac{(-1)^n \times n!}{(m+1)^{n+1}}
Let m = k, n = 1,
I = k = 0 1 ( k + 1 ) 2 = n = 1 1 n 2 \displaystyle \therefore I = - \sum_{k=0}^{\infty} -\dfrac{1}{(k+1)^{2}} = \sum_{n= 1}^{\infty} \dfrac{1}{n^{2}}
n = 1 1 n 2 \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^{2}} is the solution to the Basel Problem, and its value is equal to π 2 6 \dfrac{\pi^{2}}{6}
I = π 2 6 = 1.64493 \displaystyle \therefore I = \dfrac{\pi^{2}}{6} = 1.64493
100 I = 100 × 1.64493 = 164.493 \displaystyle \therefore 100I = 100 \times 1.64493 = 164.493



Can you please explain or prove the validity of the step in which you interchanged the order of integration and summation.

Indraneel Mukhopadhyaya - 5 years, 5 months ago

Good use of maclaurin series. try the harder version of this.

Aditya Kumar - 5 years, 4 months ago

If I am not wrong,the question asks the value of the dilogarithm function at 0,which is a standard result.

Ramiel To-ong
Jan 11, 2016

amazing solution

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