Evaluate this definite integral:
1 ln(1+x)
∫ ------ dx
0 1 + x2
where ln(n) denotes the
Natural Logarithm of n.
Note: Computer program, spreadsheet, and other numerical based methodologies are welcome, but an analytical solution is preferred.
*** Adapted from a problem which appeared at a William Lowell Putnam Competition.
Integ {x=0 to 1} ln(1+x)/(1+x^2) dx
We will start with a tangent substitution:
tan v = x
(sec v)^2 dv = dx
x=0 -> v=0
x=1 -> v=pi/4
Then after simplifying we get:
Integ {v=0 to pi/4} ln(1+tan v) dv
So now some manipulation. 1+tan v = (sin v + cos v)/(cos v). Using this plus the quotient-to-difference logarithm rule, the integral now breaks into two parts.
Integ {v=0 to pi/4} ln(sin v + cos v) dv - Integ {v=0 to pi/4} ln(cos v) dv
On the left integral we'll use the substitution:
v = pi/4 - w
dv = -dw
v=0 -> w=pi/4
v=pi/4 -> w=0
Then after simplifying and absorbing a negative to flip the bounds of integration we now have:
Integ {w=0 to pi/4} ln(sin (pi/4-w) + cos (pi/4-w)) dw
- Integ {v=0 to pi/4} ln(cos v) dv
Now we'll use basic difference formulas for sin and cos. Then the argument of ln reduces to sqrt(2)*cos w. Then after applying product-to-sum log rules we get:
Integ {w=0 to pi/4} ln(sqrt(2)) dw
+ Integ {w=0 to pi/4} ln(cos v) dw
- Integ {v=0 to pi/4} ln(cos v) dv
Of these three integrals, the last two are identical other than the variable and sign, but since these are definite integrals the variable doesn't matter so they cancel each other out.
The first integral is a trivial constant, so it evaluates to (pi/4-0)*ln(sqrt(2) = pi*ln(2)/8.
This variant is actually a very nice exercise; it is completely accessible to a high school AP calc / college Calc 2 students, but not obvious in what needs to be done.
Solution
