Integrating function containing singularity

Question

I'd like to numerically evaluate the following integral using computer software but it has a singularity at $x=1$:

\begin{equation} \int_1^{\infty} \frac{x}{1-x^4} dx \end{equation}

I was thinking of a variable transformation, rewriting the expression, or something of a kind. One of my attempts was to do differentiation under the integral sign, but I only got so far:

\begin{equation} I(b) = \int_1^{\infty} \frac{x}{1-x^4}e^{-bx} dx \end{equation}

where setting $b=0$ gives the original expression. Differentiating w.r.t. $b$ gives

\begin{equation} I'(b) = -\int_1^{\infty} \frac{x^2}{1-x^4}e^{-bx} dx = \int_1^{\infty} \frac{1}{1+x^2}e^{-bx} dx - \int_1^{\infty} \frac{1}{1-x^4}e^{-bx} dx \end{equation}

It remains to integrate the two terms w.r.t. $x$ and then $b$, but they are not standard integrals.

Is there a better, faster, or easier way that I'm not thinking of?

Answer 1

The integral is $-\infty$ because of the pole at $1$. To see this, factor

$$1 - x^4 = (1 - x)p(x)$$

where $p(x)$ is a polynomial which doesn't vanish at $1$. Then near $x = 1$ the integrand is well approximated by $$\frac 1 {p(1)} \frac 1 {1 - x}$$

which is non-integrable.

Answer 2

You can't do that, because$$\frac x{x-x^4}=\frac1{1-x}\times\frac x{1+x+x^2+x^3}$$and therefore near $1$ this functions behaves as $\frac1{4(1-x)}$. And if $a>1$, the integral$$\int_1^a\frac1{4(1-x)}\,\mathrm dx$$diverges.

Answer 3

Using partial fraction expansion reveals

$$\frac{x}{1-x^4}=\frac{x}{2(x^2+1)}-\frac{1}{4(x-1)}-\frac{1}{4(x+1)}$$

Hence, we see that for $t>1$

$$\begin{align} \int_t^\infty \frac{x}{1-x^4}\,dx&=\frac14 \log\left(\frac{1-t^2}{1+t^2}\right)\tag1 \end{align}$$

Inasmuch as $\displaystyle \lim_{t\to1^+} \log\left(\frac{1-t^2}{1+t^2}\right)$ fails to exist, the integral of interest diverges.