Best answer:
Note to poster:
This is a fun but infuriating integral. I made several attempts (double integrals, differentiating under the integral sign) on this the night before to no success. One of my attempts was very similar to John's solution, but I too made some error that I could not find. This should finally do it....
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Best answer: Note to poster:
This is a fun but infuriating integral. I made several attempts (double integrals, differentiating under the integral sign) on this the night before to no success. One of my attempts was very similar to John's solution, but I too made some error that I could not find. This should finally do it.
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First of all, we can rewrite this integral as a double integral:
∫(x = 0 to 1) (ln(1+x²)/(1+x)) dx
= ∫(x = 0 to 1) [x²/(1+x²y)]/(1+x) {for y = 0 to 1} dx
= ∫(x = 0 to 1) ∫(y = 0 to 1) [x² / ((1+x²y) (1+x))] dy dx.
Now, we reverse the order of integration:
∫(y = 0 to 1) ∫(x = 0 to 1) [x² / ((1+x²y) (1+x))] dx dy
= ∫(y = 0 to 1) ∫(x = 0 to 1) (1/(1+y)) * [(x-1)/(1+x²y) + 1/(1+x)] dx dy, by partial fractions
= ∫(y = 0 to 1) (1/(1+y)) * [∫(x = 0 to 1) (x/(1+x²y) - 1/(1+x²y) + 1/(1+x)) dx] dy
= ∫(y = 0 to 1) (1/(1+y)) [(1/(2y)) ln(1+x²y) - (1/√y) arctan(x√y) + ln(1+x)) {for x = 0 to 1}] dy
= ∫(y = 0 to 1) (1/(1+y)) [(1/(2y)) ln(1+y) - (1/√y) arctan(√y) + ln 2] dy.
Distributing:
∫(y = 0 to 1) [(1/(2y(1+y))) ln(1+y) - (1/√y) arctan(√y)/(1+y) + ln(2)/(1+y)] dy.
By partial fractions, 1/(y(y+1)) = 1/y - 1/(y+1).
So, we obtain
∫(y = 0 to 1) [(ln(1+y)/(2y) - ln(1+y)/(2(1+y))) - (1/√y) arctan(√y)/(1+y) + ln(2)/(1+y)] dy.
Integrate each term:
(i) ∫(y = 0 to 1) ln(1+y)/(2(1+y))
= ∫(u = 0 to ln 2) (1/2) u du, letting u = ln(1+y) and du = dy/(y+1)
= (1/4)(ln²2).
(ii) ∫(y = 0 to 1) (1/√y) arctan(√y) dy/(1+y)
= ∫(u = 0 to 1) 2 arctan u du/(1+u²), letting u = √y ==> y = u², dy = 2u du
= ∫(w = 0 to π/4) 2w dw, letting w = arctan u, dw = du/(1+u²).
= π²/16.
(iii) ∫(y = 0 to 1) ln(2) dy/(1+y)
= ln(2) ln(1+y) {for y = 0 to 1}
= ln²2
(iv) [The tough one...]
∫(y = 0 to 1) ln(1+y) dy/(2y)
= (1/2) ∫(y = 0 to 1) ln(1+y) dy/y
= (1/2) ∫(y = 0 to 1) [Σ(n = 1 to ∞) (-1)ⁿ⁺¹ yⁿ/n] dy/y, via power series for ln(1+y)
= (1/2) ∫(y = 0 to 1) [Σ(n = 1 to ∞) (-1)ⁿ⁺¹ yⁿ⁻¹/n] dy
= (1/2) Σ(n = 1 to ∞) (-1)ⁿ⁺¹ yⁿ/n² {for y = 0 to 1}
= (1/2) Σ(n = 1 to ∞) (-1)ⁿ⁺¹/n²
= (1/2) (π²/12); see below
= π²/24.
For the next to last step above, note that
Σ(n = 1 to ∞) (-1)ⁿ⁺¹/n²
= 1 - 1/2² + 1/3² - 1/4² + 1/5² - 1/6² + ...
= (1 + 1/2² + 1/3² + 1/4² + 1/5² + 1/6² + ...) - 2(1/2² + 1/4² + 1/6² + ...)
= (1 + 1/2² + 1/3² + ...) - 2 * (1/2²)(1 + 1/2² + 1/3² + ...)], by factoring
= (1 - 1/2)(1 + 1/2² + 1/3² + ...)
= (1/2)(π²/6), since Σ(n = 1 to ∞) 1/n² = π²/6
= π²/12.
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Putting this all together,
∫(x = 0 to 1) (ln(1+x²)/(1+x)) dx
= ∫(y = 0 to 1) [(ln(1+y)/(2y) - ln(1+y)/(2(1+y))) - (1/√y) arctan(√y)/(1+y) + ln(2)/(1+y)] dy
= π²/24 - (1/4)(ln²2) - π²/16 + ln²2
= (3/4)(ln²2) - π²/48.
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I hope this helps!
5 answers
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6 days ago