Partial fraction questions with integer areas

Question

This question is purely out of curiosity: do definite integrals that are solved by partial fractions that have integer limits of integration have non-integer areas? I know there are some integrals by parts that have integer solutions, like $$ \int_0^1 xe^x \ dx = 1$$, but partial fractions seem to have the simplest areas that still contain logarithms (at least from what I've seen at 1st year calculus level)

So I'm just wondering if there's a good example of a definite integral partial fraction that contains only integers?

Answer 1

One explicit example of such integral is $$\int_1^5 \frac{-5x^6 + 78x^5 - 245x^4 - 948x^3 + 2932x^2 + 6576x + 2016}{-x^6 + 18x^5 - 85x^4 - 60x^3 + 716x^2 + 672x}\mathop{}\!\mathrm{d}x = 20$$

Even if it may not look so at the first glance, the polynomial in the denominator is a product of six linear factors and the indefinite integral definitely (pun intended) looks ugly enough. Still, the integral over the specified interval yields an integer.

Is this the simplest possible example? Do we need to go all the way up to sixth-degree polynomials to get an integer? No, we don't:

$$\int_{-2}^2 \frac{3x^4 + 2x^3 - 8x + 12}{x^4 + 4}\mathop{}\!\mathrm{d}x=12$$

The indefinite integral still looks complicated, especially since the denominator is product of two irreducible quadratic polynomials so the partial fractions yield some inverse tangents in addition to logarithms.

But... can we do something even simpler? Yes, we can!

$$\int_{-1}^{1} \frac{2x}{x^2-4} \mathop{}\!\mathrm{d}x = 0$$

This feels a little like cheating, doesn't it? The integrand is quite clearly an odd function so, intuitively, the part on the left from zero and on the right from zero will cancel out since they have opposite signs.

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Now, what if I told you that the other two examples I provided are actually based on the same idea?

I obfuscated them slightly by adding a constant (which caused the definite integral value to be non-zero). If you look at the integrands, you'll see that the degree of the numerator and denominator is the same; so we can partially divide the two to get some integer and only deal with the remainder by the partial-fractions integration. It is precisely this some integer which provides the non-zero value of the whole integral. For example, the second integrand can be re-written as

$$\frac{3x^4 + 2x^3 - 8x + 12}{x^4 + 4} = 3 + \frac{2x^3+8x}{x^4+4}$$

and one can see that the remaining fraction is an odd function again.

My first example was additionally obfuscated by shifting the whole expression by substituting $(x-3)$ for $x$ and thus moving the interval of integration from $\langle -2,2\rangle$ to $\langle 1,5\rangle$.

Naturally, one can build infinitely many different examples by employing the same two tricks. This brings a more complicated question; one I don't know the answer for yet:

Is there a rational function with integer coefficients such that:

• its numerator is of strictly lower degree than its denominator,
• its integral over some non-trivial interval with integer bounds is an integer,
• when written as sum of partial fractions, all their denominators are polynomials irreducible over real numbers,
• is not a translate of an odd function?

Answer 2

You can make funny ones by yourself.

Consider $$I_n=\frac 1{2\pi} \int x^n\, \sqrt{\frac{4-x}{x}} \,dx$$

The antiderivative is rather intimidating $$I_n=-\frac{x^{n+1} }{n+1}\,\, _2F_1\left(-\frac{1}{2},(n+1);-n;\frac{4}{x}\right)\,i$$ $$I_n= 4^{n+1} B_{\frac{4}{x}}\left(-(n+1),\frac{3}{2}\right)\,i$$ But, integrating between $0$ and $4$, the result is simply $$J_n=\frac 1{2\pi} \int_0^4 x^n\, \sqrt{\frac{4-x}{x}} \,dx=\frac{2^{2 n}\,\, \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }_,\, \Gamma (n+2)}=C_n$$ which are Catalan numbers.

Now, make for example $$\sqrt{\frac{4-x}{x}}=P_k(t) $$ where $P_k(t)$ is a polynomial of degree $k$ in $t$ or, even worse, the ratio of polynomials. This makes $$x= \frac{4}{P_k(t)^2+1} \qquad \implies \qquad dx=-8\frac{ P_k(t) \,P_k'(t)}{\left(P_k^2(t)+1\right)^2}\,dt$$ that is to say $$I_n=\frac{2^{2 n+2}}{\pi }\int \frac{P_k^2(t)\,P_k'(t) } { \left(P_k^2(t)+1\right)^{n+2}}\,dt$$ $$I_n=-\frac{4^{n+1}}{3\pi}P_k^3(t)\,\, _2F_1\left(\frac{3}{2},n+2;\frac{5}{2};-P_k^2(t)\right)$$

I am sure that you see the monsters you can generate and the possible complexity of partial fractions.

With the corresponding bounds, you still generate Catalan numbers