I = ∫ 0 1 x − 1 ln ( x ) d x
Find 1 0 0 I upto 3 decimal places.
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Nice usage of the symmetric integration trick to simplify the working.
I
=
−
∫
0
1
1
−
x
l
n
(
x
)
d
x
Using the Maclaurin series for
1
−
x
1
∴
I
=
−
∫
0
1
l
n
(
x
)
×
(
1
+
x
²
+
x
³
.
.
.
)
d
x
=
−
∫
0
1
l
n
(
x
)
k
=
0
∑
∞
x
k
d
x
Interchanging the order of integration and summation,
∴
I
=
−
k
=
0
∑
∞
∫
0
1
x
k
l
n
(
x
)
d
x
Now, it can be proved that,
∫
0
1
x
m
l
n
n
(
x
)
d
x
=
(
m
+
1
)
n
+
1
(
−
1
)
n
×
n
!
Let m = k, n = 1,
∴
I
=
−
k
=
0
∑
∞
−
(
k
+
1
)
2
1
=
n
=
1
∑
∞
n
2
1
n
=
1
∑
∞
n
2
1
is the solution to the Basel Problem, and its value is equal to
6
π
2
∴
I
=
6
π
2
=
1
.
6
4
4
9
3
∴
1
0
0
I
=
1
0
0
×
1
.
6
4
4
9
3
=
1
6
4
.
4
9
3
Can you please explain or prove the validity of the step in which you interchanged the order of integration and summation.
Good use of maclaurin series. try the harder version of this.
If I am not wrong,the question asks the value of the dilogarithm function at 0,which is a standard result.
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Maclurin expansion of l n ( 1 + x ) = x − 2 1 x 2 + 3 1 x 3 − 4 1 x 4 + . . . where − 1 < x < 1
Replace x by x − 1 in above formula
l n ( x ) = ( x − 1 ) − 2 1 ( x − 1 ) 2 + 3 1 ( x − 1 ) 3 − 4 1 ( x − 1 ) 4 + . . .
Also by limits of integration 0 < x < 1 means − 1 < x − 1 < 0 So we can use this above formula
Now x − 1 ln x = 1 − 2 1 ( x − 1 ) + 3 1 ( x − 1 ) 2 − 4 1 ( x − 1 ) 3 + . . .
I = ∫ 0 1 x − 1 ln ( x ) d x = ∫ 0 1 ( 1 − 2 1 ( x − 1 ) + 3 1 ( x − 1 ) 2 − 4 1 ( x − 1 ) 3 + . . . ) d x
Using ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x we get
I = ∫ 0 1 ( 1 − 2 1 ( 1 − x − 1 ) + 3 1 ( 1 − x − 1 ) 2 − 4 1 ( 1 − x − 1 ) 3 + . . . ) d x
I = ∫ 0 1 ( 1 + 2 1 x + 3 1 x 2 + 4 1 x 3 + . . . ) d x
I = 1 + 2 2 1 + 3 2 1 + 4 2 1 + . . . . = 6 π 2
You can check this above sum here
1 0 0 I = 1 0 0 6 π 2 = 1 6 4 . 4 9 3