The Problem
Prove that
∫ 0 ∞ ∫ 0 ∞ e − x 2 + 4 x y + y 2 4 y log ( x y ) x y d x d y = − 2 2 π cosh − 1 ( 2 ) \int_0^\infty\!\!\int_0^\infty\frac{e^{-\frac{x^2+4xy+y^2}{4y}}\log\!\left(\frac{x}{y}\right)}{\sqrt{x}\,\sqrt{y}}\,dx\,dy \;=\; -2\sqrt{2}\,\pi\,\cosh^{-1}(2) ∫ 0 ∞ ∫ 0 ∞ x y e − 4 y x 2 + 4 x y + y 2 log ( y x ) d x d y = − 2 2 π cosh − 1 ( 2 ) This problem appeared on Mathematics Stack Exchange . The closed form involves cosh − 1 ( 2 ) = ln ( 2 + 3 ) \cosh^{-1}(2) = \ln(2 + \sqrt{3}) cosh − 1 ( 2 ) = ln ( 2 + 3 )
Solution
Step 1 — Reducing to a Single Integral
Introduce polar coordinates x = r cos θ x = r\cos\theta x = r cos θ y = r sin θ y = r\sin\theta y = r sin θ r r r
x 2 + 4 x y + y 2 4 y = r 2 ( cos 2 θ + 4 cos θ sin θ + sin 2 θ ) 4 r sin θ = r ( 1 + 4 cos θ sin θ ) 4 sin θ \frac{x^2 + 4xy + y^2}{4y} = \frac{r^2(\cos^2\theta + 4\cos\theta\sin\theta + \sin^2\theta)}{4r\sin\theta} = \frac{r(1 + 4\cos\theta\sin\theta)}{4\sin\theta} 4 y x 2 + 4 x y + y 2 = 4 r sin θ r 2 ( cos 2 θ + 4 cos θ sin θ + sin 2 θ ) = 4 sin θ r ( 1 + 4 cos θ sin θ ) and x y = r cos θ sin θ \sqrt{x}\,\sqrt{y} = r\sqrt{\cos\theta\sin\theta} x y = r cos θ sin θ log ( x / y ) = log ( cos θ / sin θ ) = log ( cot θ ) \log(x/y) = \log(\cos\theta/\sin\theta) = \log(\cot\theta) log ( x / y ) = log ( cos θ / sin θ ) = log ( cot θ )
I = ∫ 0 π / 2 ∫ 0 ∞ e − r ( 1 + 4 cos θ sin θ ) 4 sin θ log ( cot θ ) r cos θ sin θ ⋅ r d r d θ \mathcal{I} = \int_0^{\pi/2}\!\int_0^\infty \frac{e^{-\frac{r(1+4\cos\theta\sin\theta)}{4\sin\theta}}\,\log(\cot\theta)}{r\sqrt{\cos\theta\sin\theta}}\cdot r\,dr\,d\theta I = ∫ 0 π /2 ∫ 0 ∞ r cos θ sin θ e − 4 s i n θ r ( 1 + 4 c o s θ s i n θ ) log ( cot θ ) ⋅ r d r d θ The r r r ∫ 0 ∞ e − α r d r = 1 / α \int_0^\infty e^{-\alpha r}\,dr = 1/\alpha ∫ 0 ∞ e − α r d r = 1/ α α = 1 + 4 cos θ sin θ 4 sin θ \alpha = \frac{1+4\cos\theta\sin\theta}{4\sin\theta} α = 4 s i n θ 1 + 4 c o s θ s i n θ
I = ∫ 0 π / 2 4 sin θ log ( cot θ ) ( 1 + 4 cos θ sin θ ) cos θ sin θ d θ = ∫ 0 π / 2 4 tan θ log ( cot θ ) 1 + 4 cos θ sin θ d θ \mathcal{I} = \int_0^{\pi/2} \frac{4\sin\theta\,\log(\cot\theta)}{(1+4\cos\theta\sin\theta)\sqrt{\cos\theta\sin\theta}}\,d\theta = \int_0^{\pi/2}\frac{4\sqrt{\tan\theta}\;\log(\cot\theta)}{1+4\cos\theta\sin\theta}\,d\theta I = ∫ 0 π /2 ( 1 + 4 cos θ sin θ ) cos θ sin θ 4 sin θ log ( cot θ ) d θ = ∫ 0 π /2 1 + 4 cos θ sin θ 4 tan θ log ( cot θ ) d θ since sin θ cos θ sin θ = sin θ cos θ = tan θ \frac{\sin\theta}{\sqrt{\cos\theta\sin\theta}} = \sqrt{\frac{\sin\theta}{\cos\theta}} = \sqrt{\tan\theta} c o s θ s i n θ s i n θ = c o s θ s i n θ = tan θ
Step 2 — Trigonometric to Algebraic Form
Substitute y = tan θ y = \tan\theta y = tan θ d θ = d y 1 + y 2 d\theta = \frac{dy}{1+y^2} d θ = 1 + y 2 d y tan θ = y \sqrt{\tan\theta} = \sqrt{y} tan θ = y log ( cot θ ) = − ln y \log(\cot\theta) = -\ln y log ( cot θ ) = − ln y
1 + 4 cos θ sin θ = 1 + 4 y 1 + y 2 = y 2 + 4 y + 1 1 + y 2 1 + 4\cos\theta\sin\theta = 1 + \frac{4y}{1+y^2} = \frac{y^2+4y+1}{1+y^2} 1 + 4 cos θ sin θ = 1 + 1 + y 2 4 y = 1 + y 2 y 2 + 4 y + 1 Therefore
I = − 4 ∫ 0 ∞ y ln y y 2 + 4 y + 1 d y \mathcal{I} = -4\int_0^{\infty}\frac{\sqrt{y}\;\ln y}{y^2+4y+1}\,dy I = − 4 ∫ 0 ∞ y 2 + 4 y + 1 y ln y d y The roots of y 2 + 4 y + 1 = 0 y^2+4y+1=0 y 2 + 4 y + 1 = 0 y = − 2 ± 3 y=-2\pm\sqrt{3} y = − 2 ± 3 ( 0 , ∞ ) (0,\infty) ( 0 , ∞ )
Step 3 — Clearing the Square Root
Substitute y = x 2 y = x^2 y = x 2 d y = 2 x d x dy = 2x\,dx d y = 2 x d x y = x \sqrt{y}=x y = x ln y = 2 ln x \ln y = 2\ln x ln y = 2 ln x
I = − 4 ∫ 0 ∞ x ⋅ 2 ln x x 4 + 4 x 2 + 1 ⋅ 2 x d x = − 16 ∫ 0 ∞ x 2 ln x x 4 + 4 x 2 + 1 d x \mathcal{I} = -4\int_0^\infty \frac{x\cdot 2\ln x}{x^4+4x^2+1}\cdot 2x\,dx = -16\int_0^\infty \frac{x^2\ln x}{x^4+4x^2+1}\,dx I = − 4 ∫ 0 ∞ x 4 + 4 x 2 + 1 x ⋅ 2 ln x ⋅ 2 x d x = − 16 ∫ 0 ∞ x 4 + 4 x 2 + 1 x 2 ln x d x Step 4 — Integration by Parts
Factor the quartic: x 4 + 4 x 2 + 1 = ( x 2 + 2 − 3 ) ( x 2 + 2 + 3 ) x^4+4x^2+1 = (x^2+2-\sqrt{3})(x^2+2+\sqrt{3}) x 4 + 4 x 2 + 1 = ( x 2 + 2 − 3 ) ( x 2 + 2 + 3 )
We integrate by parts with u = ln x u = \ln x u = ln x d v = x 2 x 4 + 4 x 2 + 1 d x dv = \frac{x^2}{x^4+4x^2+1}\,dx d v = x 4 + 4 x 2 + 1 x 2 d x v v v
x 2 x 4 + 4 x 2 + 1 = 1 2 3 [ 2 + 3 x 2 + 2 + 3 − 2 − 3 x 2 + 2 − 3 ] \frac{x^2}{x^4+4x^2+1} = \frac{1}{2\sqrt{3}}\!\left[\frac{2+\sqrt{3}}{x^2+2+\sqrt{3}} - \frac{2-\sqrt{3}}{x^2+2-\sqrt{3}}\right] x 4 + 4 x 2 + 1 x 2 = 2 3 1 [ x 2 + 2 + 3 2 + 3 − x 2 + 2 − 3 2 − 3 ] Each term integrates to an arctangent. Rewriting in terms of arccot \operatorname{arccot} arccot [ v ln x ] 0 ∞ [v\ln x]_0^\infty [ v ln x ] 0 ∞ v = O ( 1 / x ) v = O(1/x) v = O ( 1/ x ) x → ∞ x\to\infty x → ∞ v v v x → 0 + x\to 0^+ x → 0 +
I = − 8 ∫ 0 ∞ 1 x [ 1 2 arccot ( x 2 + 1 x 2 ) + 1 6 arccot ( x 2 − 1 x 6 ) ] d x \mathcal{I} = -8\int_0^\infty \frac{1}{x}\!\left[\frac{1}{\sqrt{2}}\operatorname{arccot}\!\left(\frac{x^2+1}{x\sqrt{2}}\right) + \frac{1}{\sqrt{6}}\operatorname{arccot}\!\left(\frac{x^2-1}{x\sqrt{6}}\right)\right]dx I = − 8 ∫ 0 ∞ x 1 [ 2 1 arccot ( x 2 x 2 + 1 ) + 6 1 arccot ( x 6 x 2 − 1 ) ] d x The arguments x 2 + 1 x 2 \frac{x^2+1}{x\sqrt{2}} x 2 x 2 + 1 x 2 − 1 x 6 \frac{x^2-1}{x\sqrt{6}} x 6 x 2 − 1 1 x 2 + c \frac{1}{x^2+c} x 2 + c 1 1 ( x − a / x ) 2 + b \frac{1}{(x-a/x)^2 + b} ( x − a / x ) 2 + b 1
Step 5 — Möbius Substitution
Apply x = 1 − t 1 + t x = \frac{1-t}{1+t} x = 1 + t 1 − t d x = − 2 ( 1 + t ) 2 d t dx = -\frac{2}{(1+t)^2}\,dt d x = − ( 1 + t ) 2 2 d t
x 2 + 1 x 2 = ( 1 − t ) 2 + ( 1 + t ) 2 ( 1 − t ) ( 1 + t ) 2 = 2 1 + t 2 1 − t 2 \frac{x^2+1}{x\sqrt{2}} = \frac{(1-t)^2+(1+t)^2}{(1-t)(1+t)\sqrt{2}} = \sqrt{2}\,\frac{1+t^2}{1-t^2} x 2 x 2 + 1 = ( 1 − t ) ( 1 + t ) 2 ( 1 − t ) 2 + ( 1 + t ) 2 = 2 1 − t 2 1 + t 2 One can verify the arccot ( ⋯ / 6 ) \operatorname{arccot}(\cdots/\sqrt{6}) arccot ( ⋯ / 6 ) t ↦ − t t \mapsto -t t ↦ − t
I = − 16 2 ∫ 0 1 1 1 − t 2 arccot ( 2 1 + t 2 1 − t 2 ) d t \mathcal{I} = -16\sqrt{2}\int_0^1 \frac{1}{1-t^2}\operatorname{arccot}\!\left(\sqrt{2}\,\frac{1+t^2}{1-t^2}\right)dt I = − 16 2 ∫ 0 1 1 − t 2 1 arccot ( 2 1 − t 2 1 + t 2 ) d t Step 6 — Evaluating the Final Integral
Substitute t = tanh ( u / 2 ) t = \tanh(u/2) t = tanh ( u /2 ) 1 − t 2 = sech 2 ( u / 2 ) ⋅ 2 1 + cosh u 1-t^2 = \operatorname{sech}^2(u/2)\cdot\frac{2}{1+\cosh u} 1 − t 2 = sech 2 ( u /2 ) ⋅ 1 + c o s h u 2
∫ 0 1 1 1 − t 2 arccot ( a 1 + t 2 1 − t 2 ) d t = π 8 cosh − 1 ( 2 a 2 + 1 ) \int_0^1\frac{1}{1-t^2}\operatorname{arccot}\!\left(a\,\frac{1+t^2}{1-t^2}\right)dt = \frac{\pi}{8}\cosh^{-1}(2a^2+1) ∫ 0 1 1 − t 2 1 arccot ( a 1 − t 2 1 + t 2 ) d t = 8 π cosh − 1 ( 2 a 2 + 1 ) Setting a = 2 a = \sqrt{2} a = 2 cosh − 1 ( 2 ⋅ 2 + 1 ) = cosh − 1 ( 5 ) \cosh^{-1}(2\cdot 2+1) = \cosh^{-1}(5) cosh − 1 ( 2 ⋅ 2 + 1 ) = cosh − 1 ( 5 ) 6 \sqrt{6} 6
− 16 2 ⋅ π 8 ⋅ cosh − 1 ( 2 ) = − 2 2 π cosh − 1 ( 2 ) -16\sqrt{2}\cdot\frac{\pi}{8}\cdot\cosh^{-1}(2) = -2\sqrt{2}\,\pi\,\cosh^{-1}(2) − 16 2 ⋅ 8 π ⋅ cosh − 1 ( 2 ) = − 2 2 π cosh − 1 ( 2 ) Final Result
∫ 0 ∞ ∫ 0 ∞ e − x 2 + 4 x y + y 2 4 y log ( x y ) x y d x d y = − 2 2 π cosh − 1 ( 2 ) \boxed{\displaystyle\int_0^\infty\!\!\int_0^\infty\frac{e^{-\frac{x^2+4xy+y^2}{4y}}\log\!\left(\frac{x}{y}\right)}{\sqrt{x}\,\sqrt{y}}\,dx\,dy \;=\; -2\sqrt{2}\,\pi\,\cosh^{-1}(2)} ∫ 0 ∞ ∫ 0 ∞ x y e − 4 y x 2 + 4 x y + y 2 log ( y x ) d x d y = − 2 2 π cosh − 1 ( 2 ) where cosh − 1 ( 2 ) = ln ( 2 + 3 ) ≈ 1.317 \cosh^{-1}(2) = \ln(2+\sqrt{3})\approx 1.317 cosh − 1 ( 2 ) = ln ( 2 + 3 ) ≈ 1.317 I ≈ − 11.699 \mathcal{I}\approx -11.699 I ≈ − 11.699
Complete Solution at a Glance
I = ∫ 0 π / 2 4 tan θ log ( cot θ ) 1 + 4 cos θ sin θ d θ = y = tan θ − 4 ∫ 0 ∞ y ln y y 2 + 4 y + 1 d y = y = x 2 − 16 ∫ 0 ∞ x 2 ln x x 4 + 4 x 2 + 1 d x = IBP − 8 ∫ 0 ∞ 1 x [ 1 2 arccot ( x 2 + 1 x 2 ) + 1 6 arccot ( x 2 − 1 x 6 ) ] d x = x = 1 − t 1 + t − 16 2 ∫ 0 1 1 1 − t 2 arccot ( 2 1 + t 2 1 − t 2 ) d t = − 2 2 π cosh − 1 ( 2 ) \begin{aligned}
\mathcal{I} &= \int_0^{\pi/2} \frac{4\sqrt{\tan\theta}\;\log(\cot\theta)}{1 + 4\cos\theta\sin\theta}\,d\theta \\[8pt]
&\overset{y\,=\,\tan\theta}{=}\; -4\int_0^{\infty} \frac{\sqrt{y}\;\ln y}{y^2+4y+1}\,dy \\[8pt]
&\overset{y\,=\,x^2}{=}\; -16\int_0^{\infty}\frac{x^2\ln x}{x^4+4x^2+1}\,dx \\[8pt]
&\overset{\text{IBP}}{=}\; -8\int_0^{\infty}\frac{1}{x}\!\left[\frac{1}{\sqrt{2}}\operatorname{arccot}\!\left(\frac{x^2+1}{x\sqrt{2}}\right)+\frac{1}{\sqrt{6}}\operatorname{arccot}\!\left(\frac{x^2-1}{x\sqrt{6}}\right)\right]dx \\[8pt]
&\overset{x\,=\,\frac{1-t}{1+t}}{=}\; -16\sqrt{2}\int_0^1 \frac{1}{1-t^2}\operatorname{arccot}\!\left(\sqrt{2}\,\frac{1+t^2}{1-t^2}\right)dt \\[8pt]
&=\; -2\sqrt{2}\,\pi\,\cosh^{-1}(2)
\end{aligned} I = ∫ 0 π /2 1 + 4 cos θ sin θ 4 tan θ log ( cot θ ) d θ = y = t a n θ − 4 ∫ 0 ∞ y 2 + 4 y + 1 y ln y d y = y = x 2 − 16 ∫ 0 ∞ x 4 + 4 x 2 + 1 x 2 ln x d x = IBP − 8 ∫ 0 ∞ x 1 [ 2 1 arccot ( x 2 x 2 + 1 ) + 6 1 arccot ( x 6 x 2 − 1 ) ] d x = x = 1 + t 1 − t − 16 2 ∫ 0 1 1 − t 2 1 arccot ( 2 1 − t 2 1 + t 2 ) d t = − 2 2 π cosh − 1 ( 2 ) References
Original problem and solution on Mathematics Stack Exchange
Gradshteyn, I. S. & Ryzhik, I. M., Table of Integrals, Series, and Products , 8th edition, Academic Press, 2014.
Boros, G. & Moll, V. H., Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals , Cambridge University Press, 2004.