Really? Any cubic polynomial?

Calculus Level 4

True or False?

Let $P(x)$ be a cubic polynomial on $x$ with real coefficients. Then $\int \frac{1}{P(x)} \, dx$ has a closed form .

True False

1 solution

Shaun Leong
Aug 6, 2016

Let $P(x)=k(x-a)(x-b)(x-c)$

To find the desired integral, we use partial fractions .

$\int \frac{1}{P(x)} dx$ $=\frac1k \int \left(\frac{1}{(a-b)(a-c)(x-a)}+\frac{1}{(b-a)(b-c)(x-b)}+\frac{1}{(c-a)(c-b)(x-c)}\right) dx$ $=\frac1k \left(\frac{\ln |x-a|}{(a-b)(a-c)}+\frac{\ln |x-b|}{(b-a)(b-c)}+\frac{\ln |x-c|}{(c-a)(c-b)}\right) + \mbox{const.}$

Thus the answer is $\boxed{\mbox{Yes}}$ .

I didn't state that all the roots are real. What happens if there are two complex roots? How would you evaluate $\ln | x - \text{some complex number} |$ ?

Pi Han Goh - 4 years, 10 months ago

Hmm... I didn't consider that. Maybe even though each term is complex the indefinite integral remains real?

With complex roots I'm not sure how the real coefficients condition comes in though.

Shaun Leong - 4 years, 10 months ago

Hint : Think of hyperbolic functions (or maybe inverse hyperbolic functions?).

On the other hand, can you prove that the denominator always has a real root, so that you will (eventually) left with a quadratic polynomial?

Pi Han Goh - 4 years, 10 months ago

@Pi Han Goh – A cubic polynomial will always intersect the x-axis, so it has a real root. Instead of 3 linear fractions, we can get a linear and a quadratic, and then use arctan (arctanh?) to integrate.

Shaun Leong - 4 years, 10 months ago

@Shaun Leong – Yeah, but how do you know it actually intersect the $x$ -axis? It seems obvious right? But how do you prove it?

Pi Han Goh - 4 years, 10 months ago

@Pi Han Goh – If the leading coefficient is positive then as x tends to negative infinity the expression tend to negative infinity. As x tends to infinity the expression tends to infinity.

Vice versa for a negative leading coefficient.

Shaun Leong - 4 years, 10 months ago

@Shaun Leong – Great. That's intermediate value theorem.

Now, since you have shown that there is a real root, can you prove that this root has a "closed form"?

Pi Han Goh - 4 years, 10 months ago

Really? Any cubic polynomial?

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