Integrate It to Infinity

Calculus Level 5

∫ 0 ∞ x 2 ( x 2 + a 2 ) ( x 2 + b 2 ) ( x 2 + c 2 ) d x = π n \large \int_0^\infty\dfrac{x^2 }{(x^2+a^2)(x^2+b^2)(x^2+c^2)} \, dx =\dfrac{\pi}{n}

Let a , b a,b and c c be the real roots of the cubic equation x 3 − 7 x 2 + 10 x − 2 = 0 x^3-7x^2+10x-2=0 . Find the value of the positive integer n n that satisfy the equation above.


The answer is 136.

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Russelle Guadalupe by Russelle Guadalupe , 25, Philippines

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2 solutions

Otto Bretscher
Dec 2, 2015

Brief outline of a solution: The integral comes out to be π 2 ( a + b ) ( a + c ) ( b + c ) = 1 2 π ( a + b + c ) ( a b + a c + b c ) − a b c = π 2 ( 7 ∗ 10 − 2 ) \frac{\pi}{2(a+b)(a+c)(b+c)}=\frac{1}{2}\frac{\pi}{(a+b+c)(ab+ac+bc)-abc}=\frac{\pi}{2(7*10-2)} by Viete, so that the denominator we seek is 136 \boxed{136}

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For those who want to see a detailed solution:

This question involves Integration with Partial Fractions . Let A , B , C , D , E A,B,C,D,E and F F be constants such that

x 2 ( x 2 + a 2 ) ( x 2 + b 2 ) ( x 2 + c 2 ) ≡ A x + B x 2 + a 2 + C x + D x 2 + b 2 + E x + F x 2 + c 2 . \dfrac{x^2}{(x^2+a^2)(x^2+b^2)(x^2+c^2)} \equiv \dfrac {Ax+B}{x^2 + a^2} + \dfrac{Cx + D}{x^2 + b^2} + \dfrac{Ex+F}{x^2 + c^2} .

Note that g ( x ) = x 2 ( x 2 + a 2 ) ( x 2 + b 2 ) ( x 2 + c 2 ) g(x) = \dfrac{x^2}{(x^2+a^2)(x^2+b^2)(x^2+c^2)} is an even functions, thus the coefficients of all the odd powers of x x must be equal to 0, thus A = C = E = 0 A=C=E=0 , and we are left with

x 2 ( x 2 + a 2 ) ( x 2 + b 2 ) ( x 2 + c 2 ) ≡ B x 2 + a 2 + D x 2 + b 2 + F x 2 + c 2 . \dfrac{x^2}{(x^2+a^2)(x^2+b^2)(x^2+c^2)} \equiv \dfrac {B}{x^2 + a^2} + \dfrac{ D}{x^2 + b^2} + \dfrac{F}{x^2 + c^2} .

By Partial Fractions - Cover Up Rule ,

B = − a 2 ( a 2 − b 2 ) ( a 2 − c 2 ) , D = − b 2 ( b 2 − c 2 ) ( b 2 − a 2 ) , F = − c 2 ( c 2 − a 2 ) ( c 2 − b 2 ) . B =- \dfrac{a^2}{(a^2-b^2)(a^2-c^2)} , D = -\dfrac{b^2}{(b^2-c^2)(b^2-a^2)} , F = -\dfrac{c^2}{(c^2-a^2)(c^2-b^2) } .

Thus, the integral in question is equal to

∫ 0 ∞ [ − a 2 ( a 2 − b 2 ) ( a 2 − c 2 ) ⋅ 1 x 2 + a 2 − b 2 ( b 2 − c 2 ) ( b 2 − a 2 ) ⋅ 1 x 2 + b 2 − c 2 ( c 2 − a 2 ) ( c 2 − b 2 ) ⋅ 1 x 2 + c 2 ] d x \int_0^\infty \left [ - \dfrac{a^2}{(a^2-b^2)(a^2-c^2)} \cdot \dfrac1{x^2+a^2} - \dfrac{b^2}{(b^2-c^2)(b^2-a^2)} \cdot \dfrac1{x^2+b^2} - \dfrac{c^2}{(c^2-a^2)(c^2-b^2)} \cdot \dfrac1{x^2+c^2} \right ] \, dx

What's left to solve is to determine the integrals in the form of ∫ 0 ∞ 1 x 2 + a 2 d x \displaystyle \int_0^\infty \dfrac1{x^2+a^2} \, dx , we apply the Integration U-substitution - Trigonometric with x = a tan ⁡ θ x = a \tan \theta to get ∫ 0 ∞ 1 x 2 + a 2 d x = 1 a tan ⁡ − 1 ( x a ) ∣ x = 0 x → ∞ = π 2 a \displaystyle \int_0^\infty \dfrac1{x^2+a^2} \, dx = \dfrac1a \left . \tan^{-1} \left( \dfrac xa \right) \right |_{x=0}^{x\to \infty} = \dfrac{ \pi}{2a} .

Substitute them and simplify,

π 2 ( a ( c 2 − b 2 ) + b ( a 2 − c 2 ) + c ( b 2 − a 2 ) ( a 2 − b 2 ) ( a 2 − c 2 ) ( b 2 − c 2 ) ) = π 2 ( a + b ) ( a + c ) ( b + c ) . \dfrac \pi 2 \left ( \dfrac{a(c^2-b^2)+b(a^2-c^2)+c(b^2-a^2)}{(a^2-b^2)(a^2-c^2)(b^2-c^2)}\right) = \dfrac{\pi}{2(a+b)(a+c)(b+c)} .

What's left to do is to compute N = 2 ( a + b ) ( a + c ) ( b + c ) N=2(a+b)(a+c)(b+c) .

By Vieta's Formula - Higher Degrees , a + b + c = 7 a+b+c = 7 , a b + b c + a c = 10 ab+bc+ac = 10 , a b c = 2 abc= 2 .

And we apply the algebraic identity,

( a + b ) ( a + c ) ( b + c ) = ( a + b + c ) ( a b + b c + a c ) − a b c = 7 ( 10 ) − 2 = 68. (a+b)(a+c)(b+c) = (a+b+c)(ab+bc+ac) - abc = 7(10) - 2 = 68.

Hence, N = 2 ( 68 ) = 136 N = 2(68) = \boxed{136} .

Pi Han Goh - 5 years, 6 months ago

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This is a very patient and detailed solution...thank you!

Your coefficients B , D , E B,D,E are not quite correct (see my solution)... this will affect your computations later on.

When I have a bit of free time, I will try to write my solution a bit more coherently.

Otto Bretscher - 5 years, 6 months ago

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Okay fixed thanks

Pi Han Goh - 5 years, 6 months ago

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@Pi Han Goh – Almost fixed... B B does not contain the factor x 2 + a 2 x^2+a^2 etc.

Otto Bretscher - 5 years, 6 months ago

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@Otto Bretscher – K thanks againn

Pi Han Goh - 5 years, 6 months ago

The integral comes out to be π 2 ( a + b ) ( a + c ) ( b + c ) \frac{\pi}{2(a+b)(a+c)(b+c)}

How is this so?

Pi Han Goh - 5 years, 6 months ago

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I used partial fraction to write a cyclic sum − a 2 ( a 2 − b 2 ) ( a 2 − c 2 ) ( a 2 + x 2 ) -\frac{a^2}{(a^2-b^2)(a^2-c^2)(a^2+x^2)} , used the arctan to find ∫ 0 ∞ 1 x 2 + a 2 d x = π 2 a \int_{0}^{\infty}\frac{1}{x^2+a^2}dx=\frac{\pi}{2a} for positive a a , and then simplified.

Otto Bretscher - 5 years, 6 months ago

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You're still missing a lot of crucial steps. It's hard to understand what you're saying.

Pi Han Goh - 5 years, 6 months ago

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@Pi Han Goh – I think you are just giving me a hard time, Comrade Pi Han Goh ;) The "crucial steps" I'm "missing" are just basic algebra. Which step(s) do you want me to elaborate on : (a) The partial fractions at the beginning (b) the integration (c) adding the three integrals up to get π 2 ( a + b ) ( a + c ) ( b + c ) \frac{\pi}{2(a+b)(a+c)(b+c)} ?

Otto Bretscher - 5 years, 6 months ago

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@Otto Bretscher – Ideally, all three. Keep in mind that not all users are Level 10 calculus users like you, so it's not likely most people can understand what you're saying.

Pi Han Goh - 5 years, 6 months ago

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@Pi Han Goh – (a) Write the integrand as A x 2 + a 2 + B x 2 + b 2 + C x 2 + c 2 \frac{A}{x^2+a^2}+\frac{B}{x^2+b^2}+\frac{C}{x^2+c^2} . Multiplying through with ( x 2 + a 2 ) ( x 2 + b 2 ) ( x 2 + c 2 ) (x^2+a^2)(x^2+b^2)(x^2+c^2) , we see that it is required that x 2 = A ( x 2 + b 2 ) ( x 2 + c 2 ) + ( x 2 + a 2 ) ( . . . ) x^2=A(x^2+b^2)(x^2+c^2)+(x^2+a^2)(...) . Letting x 2 = − a 2 x^2=-a^2 gives A = − a 2 ( a 2 − b 2 ) ( a 2 − c 2 ) A=-\frac{a^2}{(a^2-b^2)(a^2-c^2)} ... find B B and C C analogously.

(b) ∫ 0 ∞ 1 x 2 + a 2 d x = 1 a [ arctan ⁡ ( x / a ) ] 0 ∞ = π 2 a \int_{0}^{\infty}\frac{1}{x^2+a^2}dx=\frac{1}{a}[\arctan(x/a)]_{0}^{\infty}=\frac{\pi}{2a}

(c) π 2 a ( c 2 − b 2 ) + b ( a 2 − c 2 ) + c ( b 2 − a 2 ) ( a 2 − b 2 ) ( a 2 − c 2 ) ( b 2 − c 2 ) = π 2 ( a + b ) ( a + c ) ( b + c ) \frac{\pi}{2}\frac{a(c^2-b^2)+b(a^2-c^2)+c(b^2-a^2)}{(a^2-b^2)(a^2-c^2)(b^2-c^2)}=\frac{\pi}{2(a+b)(a+c)(b+c)}

Otto Bretscher - 5 years, 6 months ago

@Pi Han Goh – Is there level 10 ??

Shyambhu Mukherjee - 5 years, 6 months ago

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@Shyambhu Mukherjee – No. I was being sarcastic. I told the person who posted this solution that not everyone is not as great as him, and most people can't understand what he's saying. So if there is a level 10 on Brilliant, then he will definitely qualify as level 10 calculus user, while the rest of us are level 5 or lower.

Pi Han Goh - 5 years, 6 months ago

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@Pi Han Goh – Oho I thought B^2 has likely levels.

Shyambhu Mukherjee - 5 years, 6 months ago

will u please tell me how the integral comes out ??

Swarnendu Bhattacharjee - 5 years, 6 months ago

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Did not u solve the question??

Shyambhu Mukherjee - 5 years, 6 months ago
Noam Pirani
Apr 21, 2016

I used a method for computing improper integrals by complex analysis. Noticing that the integrand is an even function and denoting the intrgral by I and the integrand by F, we see that 2 I = ∫ − ∞ ∞ F ( x ) 2I=\int_{-\infty}^{\infty}F(x) . Now we look at the integrand as a function of complex veriable, and integrate it along the upper half circle with radius r, denoted C r C_r . the function is meromorphic so we now that if r is large enough, the whole integral will be equal to the residues of F(z) in the upper half plane, namely { a i , b i , c i } \{ai,bi,ci\} , times 2 π i 2\pi i . Thus:

∫ C r F ( z ) = 2 π i [ R e s z = a i ( F ( z ) ) + R e s z = b i ( F ( z ) ) + R e s z = c i ( F ( z ) ) ] \int_{C_r}F(z)=2\pi i \left[ Res_{z=ai}(F(z))+Res_{z=bi}(F(z))+Res_{z=ci}(F(z)) \right]

But from Jordan's lemma we know that as r → ∞ r\to\infty , the LHS of the equation is just 2 I 2I . Thus:

2 I π = 2 i [ R e s z = a i ( F ( z ) ) + R e s z = b i ( F ( z ) ) + R e s z = c i ( F ( z ) ) ] = 2 i [ − a 2 ( 2 a i ( b 2 − a 2 ) ( c 2 − a 2 ) + − b 2 2 b i ( a 2 − b 2 ) ( c 2 − b 2 ) + − c 2 2 c i ( a 2 − c 2 ) ( b 2 − c 2 ) ] = ∑ c y c ( a 2 b − a 2 c ) ∑ c y c ( a 4 b 2 − a 4 c 2 ) \frac{2I}{\pi}=2 i\left[ Res_{z=ai}(F(z))+Res_{z=bi}(F(z))+Res_{z=ci}(F(z)) \right]= \\ 2 i \left[ \frac {-a^2}{(2ai(b^2-a^2)(c^2-a^2)}+\frac {-b^2}{2bi(a^2-b^2)(c^2-b^2)}+\frac {-c^2}{2ci(a^2-c^2)(b^2-c^2)}\right] = \frac{\displaystyle\sum_{cyc}{(a^2 b-a^2 c)}}{\displaystyle\sum_{cyc}{(a^4 b^2-a^4 c^2)}}

Denoting M = a + b + c , N = a b + a c + b c , P = a b c M=a+b+c,N=ab+ac+bc,P=abc we have that the RHS of the equation is-

( a − b ) ( a − c ) ( b − c ) ( a − b ) ( a − c ) ( b − c ) ( M N − P ) \frac{(a-b)(a-c)(b-c)}{(a-b)(a-c)(b-c)(MN-P)} which means that I = 1 2 ( M N − P ) I=\frac{1}{2(MN-P)} . From Vieta's formula we get the answer.

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