Unique Triple Integral

Calculus Level 5

∫ 0 1 ∫ 0 1 ∫ 0 1 ( x y z ) x y z d x d y d z = ? \displaystyle \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \left ( xyz \right )^{xyz}\,dx \, dy\, dz = \, ?

Give your answer to 3 decimal places.


The answer is 0.83493.

Uzumaki Nagato Tenshou Uzumaki by uzumaki nagato tenshou u… , 26

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3 solutions

Mark Hennings
Jan 21, 2017

We have I = ∫ 0 1 ∫ 0 1 ∫ 0 1 ( x y z ) x y z d x d y d z = ∑ n = 0 ∞ 1 n ! ∫ 0 1 ∫ 0 1 ∫ 0 1 ( x y z ln ⁡ ( x y z ) ) n d x d y d z = ∑ a , b , c = 0 ∞ 1 a ! b ! c ! ∫ 0 1 ∫ 0 1 ∫ 0 1 x a + b + c y a + b + c z a + b + c ( ln ⁡ x ) a ( ln ⁡ y ) b ( ln ⁡ z ) c d x d y d z = ∑ a , b , c = 0 ∞ 1 a ! b ! c ! ∫ 0 ! x a + b + c ( ln ⁡ x ) a d x ∫ 0 1 y a + b + c ( ln ⁡ y ) b d y ∫ 0 1 z a + b + c ( ln ⁡ z ) c d z = ∑ a , b , c = 0 ∞ 1 a ! b ! c ! × ( − 1 ) a a ! ( a + b + c + 1 ) a + 1 × ( − 1 ) b b ! ( a + b + c + 1 ) b + 1 × ( − 1 ) c c ! ( a + b + c + 1 ) c + 1 = ∑ a , b , c = 0 ∞ ( − 1 ) a + b + c ( a + b + c + 1 ) a + b + c + 3 = ∑ N = 0 ∞ ( − 1 ) N ( N + 1 ) N + 3 × 1 2 ( N + 1 ) ( N + 2 ) = 1 2 ∑ N = 0 ∞ ( − 1 ) N ( N + 2 ) ( N + 1 ) N + 2 = 0.83493 \begin{aligned} I & = \int_0^1 \int_0^1 \int_0^1 (xyz)^{xyz}\,dx\,dy\,dz \; = \; \sum_{n=0}^\infty \frac{1}{n!}\int_0^1 \int_0^1 \int_0^1 \big(xyz \ln(xyz)\big)^n\,dx\,dy\,dz \\ & = \sum_{a,b,c=0}^\infty \frac{1}{a!b!c!}\int_0^1\int_0^1\int_0^1 x^{a+b+c}y^{a+b+c}z^{a+b+c} (\ln x)^a (\ln y)^b (\ln z)^c \,dx\,dy\,dz \\ & = \sum_{a,b,c=0}^\infty \frac{1}{a!b!c!}\int_0^! x^{a+b+c}(\ln x)^a\,dx \int_0^1 y^{a+b+c}(\ln y)^b\,dy \int_0^1 z^{a+b+c}(\ln z)^c\,dz \\ & = \sum_{a,b,c=0}^\infty \frac{1}{a!b!c!} \times \frac{(-1)^a a!}{(a+b+c+1)^{a+1}} \times \frac{(-1)^b b!}{(a+b+c+1)^{b+1}} \times \frac{(-1)^c c!}{(a+b+c+1)^{c+1}} \\ & = \sum_{a,b,c=0}^\infty \frac{(-1)^{a+b+c}}{(a+b+c+1)^{a+b+c+3}} \; = \; \sum_{N=0}^\infty \frac{(-1)^N}{(N+1)^{N+3}} \times \tfrac12(N+1)(N+2) \\ & = \tfrac12\sum_{N=0}^\infty \frac{(-1)^N(N+2)}{(N+1)^{N+2}} \; = \; \boxed{0.83493} \end{aligned}

16 Helpful 1 Interesting 2 Brilliant 1 Confused

Can u explain 2nd line

Kushal Bose - 4 years, 4 months ago

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I am applying the multinomial expansion ( X + Y + Z ) n = ∑ a , b , c ≥ 0 a + b + c = n n ! a ! b ! c ! X a Y b Z c (X + Y + Z)^n \; = \; \sum_{{a,b,c \ge 0} \atop {a+b+c=n}} \frac{n!}{a!b!c!}X^aY^bZ^c to the term ( ln ⁡ ( x y z ) ) n = ( ln ⁡ x + ln ⁡ y + ln ⁡ z ) n \big(\ln(xyz)\big)^n \; = \; (\ln x + \ln y + \ln z)^n

Mark Hennings - 4 years, 4 months ago

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ok i got it

Kushal Bose - 4 years, 4 months ago

But how to calculate last sequence without calculator or any else?

uzumaki nagato tenshou uzumaki - 4 years, 4 months ago

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Well, I don't think the answer is rational, and therefore any calculation, even from a formula, will require a calculator.

If you have a definite value of this Sophomore's Dream-like sum, post it...

Of course, if you want an exact solution to a problem, post it in a way that requires an exact solution. WA can evaluate this integral numerically. I took the time to express it as a highly convergent sum, so that not many terms are needed to obtain 3DP.

Mark Hennings - 4 years, 4 months ago

How did you get from the third to the fourth line? Is this an well known identity?

Chaebum Sheen - 4 years, 3 months ago

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To integrate ∫ 0 1 x m ( ln ⁡ x ) m d x \int_0^1 x^m (\ln x)^m\,dx try integration by parts...

Mark Hennings - 4 years, 3 months ago

or u can use Gamma Function ∫ 0 1 x m ( ln ⁡ x ) n d x \displaystyle \int_0^1 x^{m}(\ln x)^{n} dx let u = ln ⁡ x u=\ln x then ∫ − ∞ 0 e ( m + 1 ) u u n d u \displaystyle \int_{-\infty}^{0} e^{(m+1)u} u^{n} du let again − t = ( m + 1 ) u -t=(m+1)u then − 1 m + 1 ∫ ∞ 0 e − t ( − t m + 1 ) n d t \displaystyle -\dfrac{1}{m+1} \int_{\infty}^{0} e^{-t} \left( \dfrac{-t}{m+1}\right)^{n} dt = ( − 1 ) n ( 1 m + 1 ) n + 1 ∫ 0 ∞ e − t t n d t = \displaystyle (-1)^{n}\left( \dfrac{1}{m+1}\right)^{n+1} \int_{0}^{\infty} e^{-t} t^{n} dt = ( − 1 ) n ( m + 1 ) n + 1 Γ ( n + 1 ) = \displaystyle \dfrac{(-1)^{n}}{(m+1)^{n+1}} \Gamma{(n+1)} = ( − 1 ) n n ! ( m + 1 ) n + 1 = \displaystyle \dfrac{(-1)^{n}n!}{(m+1)^{n+1}}

uzumaki nagato tenshou uzumaki - 4 years, 3 months ago

Every time I see a great solution by you, I am blown away! You have got to be the smartest mathematician I have ever seen at work. (Just so you know, I just used Wolfram Alpha to approximate the integral for me. Let's just say it replaces another one that I should've gotten right.) I thought to use e e , but I always forget how helpful using series can be for integration. But even if I had really tried hard on this problem and thought to use series, I probably would have just assumed the approach wouldn't have worked here and given up. Sometimes I wonder if you are only one person or a mathematician army.

James Wilson - 3 years, 5 months ago

Which of your solutions is your favorite?

James Wilson - 3 years, 5 months ago

You lost me right at the first step.

Dhvanit Beniwal - 2 years, 11 months ago

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I am writing ( x y z ) x y z = e x y z ln ⁡ ( x y z ) (xyz)^{xyz}=e^{xyz\ln(xyz)} , and expanding the exponential.

Mark Hennings - 2 years, 11 months ago
Leonel Castillo
Aug 26, 2018

To approximate the solution first we reduce it to a single integral: Let u = x y u = xy , then d u = y d x du = y dx and thus ∫ 0 1 ∫ 0 1 ∫ 0 1 ( x y z ) x y z d x d y d z = ∫ 0 1 ∫ 0 1 ∫ 0 y ( u z ) u z y d u d y d z = ∫ 0 1 ∫ 0 1 ∫ u 1 ( u z ) u z y d y d u d z = ∫ 0 1 ∫ 0 1 log ⁡ 1 u ( u z ) u z d u d z \int_0^1 \int_0^1 \int_0^1 (xyz)^{xyz} dx dy dz = \int_0^1 \int _0^1 \int_0^y \frac{(uz)^{uz}}{y} du dy dz = \int_0^1 \int_0^1 \int_u^1 \frac{(uz)^{uz}}{y} dy du dz = \int_0^1 \int_0^1 \log \frac{1}{u} (uz)^{uz} du dz Similarly, let's take w = u z ⟹ d w = z d u w = uz \implies dw = zdu ∫ 0 1 ∫ 0 1 log ⁡ 1 u ( u z ) u z d u d z = ∫ 0 1 ∫ 0 z log ⁡ z w w w z d w d z = ∫ 0 1 ∫ w 1 log ⁡ z w w w z d z d w = 1 2 ∫ 0 1 w w log ⁡ ( w ) 2 d w \int_0^1 \int_0^1 \log \frac{1}{u} (uz)^{uz} du dz = \int_0^1 \int_0^z \frac{\log \frac{z}{w} w^w}{z} dw dz = \int_0^1 \int_w^1 \frac{\log \frac{z}{w} w^w}{z} dz dw = \frac{1}{2} \int_0^1 w^w \log(w)^2 dw

So the triple integral is equal to the single integral 1 2 ∫ 0 1 x x log ⁡ ( x ) 2 d x \frac{1}{2} \int_0^1 x^x \log(x)^2 dx . We can also find a series expansion through standard methods:

1 2 ∫ 0 1 x x log ⁡ ( x ) 2 d x = 1 2 ∫ 0 1 e x log ⁡ x log ⁡ ( x ) 2 d x = 1 2 ∫ 0 1 ∑ n = 0 ∞ x n log ⁡ ( x ) n + 2 n ! d x = 1 2 ∑ n = 0 ∞ 1 n ! ∫ 0 1 x n log ⁡ ( x ) n + 2 d x = 1 2 ∑ n = 0 ∞ 1 n ! ( − 1 ) n + 2 ( n + 2 ) ! ( n + 1 ) n + 3 \frac{1}{2} \int_0^1 x^x \log(x)^2 dx = \frac{1}{2} \int_0^1 e^{x \log x} \log(x)^2 dx = \frac{1}{2} \int_0^1 \sum_{n=0}^{\infty} \frac{x^n \log(x)^{n+2}}{n!} dx = \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{n!} \int_0^1 x^n \log(x)^{n+2} dx = \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{n!} \frac{(-1)^{n+2} (n+2)!}{(n+1)^{n+3}} = 1 2 ∑ n = 0 ∞ ( − 1 ) n ( n + 1 ) ( n + 2 ) ( n + 1 ) n + 3 = 1 2 ∑ n = 0 ∞ ( − 1 ) n ( n + 2 ) ( n + 1 ) n + 2 = \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n (n+1)(n+2)}{(n+1)^{n+3}} = \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n (n+2)}{(n+1)^{n+2}}

Thus: ∫ 0 1 ∫ 0 1 ∫ 0 1 ( x y z ) x y z d x d y d z = 1 2 ∫ 0 1 x x log ⁡ ( x ) 2 d x = 1 2 ∑ n = 0 ∞ ( − 1 ) n ( n + 2 ) ( n + 1 ) n + 2 \int_0^1 \int_0^1 \int_0^1 (xyz)^{xyz} dx dy dz = \frac{1}{2} \int_0^1 x^x \log(x)^2 dx = \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n (n+2)}{(n+1)^{n+2}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Vinod Kumar
Jun 25, 2018

Used Wolfram alpha 3d integration to get the desired answer = 0.83493

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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