Is the integral $$\int^1_{-1} \left(\frac 1 {x^5} + x^2\right)\ dx$$ convergent or divergent?

My working

Clearly, $$\int^1_{-1} x^2\ dx$$ is always convergent, so we check $$\int^1_{-1} \frac 1 {x^5}\ dx,$$ where $$\begin{aligned} \int^1_{-1} \frac 1 {x^5}\ dx & = \int^0_{-1} \frac 1 {x^5}\ dx + \int^1_0 \frac 1 {x^5}\ dx\\[2 mm] & = \lim\limits_{c \to 0^-} \left[-\frac 1 {4x^4}\right]^c_{-1} + \lim\limits_{c \to 0^+} \left[-\frac 1 {4x^4}\right]^1_c\\[2 mm] & = \lim\limits_{c \to 0^-} \left(-\frac 1 {4c^4}\right) + \lim\limits_{c \to 0^+} \left(\frac 1 {4c^4}\right)\\[2 mm] & = -\infty + \infty, \end{aligned}$$ which is indeterminate, so I concluded that the integral is divergent. However, the solution goes from $$\lim\limits_{c \to 0^-} \left(-\frac 1 {4c^4}\right) + \lim\limits_{c \to 0^+} \left(\frac 1 {4c^4}\right)$$ to $0$ and concludes that the integral is convergent.

I am not sure where I have gone wrong, so any intuitive explanations will be greatly appreciated!

  • 1
    $\begingroup$ Consider $$f:[-1,1]\to\Bbb R \\ f(x)=\begin{cases}1/x & x>0 \\ 0 & x\leq 0\end{cases}$$ It is very clear (exercise!) that $|f(x)|\leq \left|\frac{1}{x^5}+x^2\right|$. It is also very clear that $\int_{-1}^1 f(x)\mathrm dx$ is divergent. Hence $\int_{-1}^1 (\frac{1}{x^5}+x^2)~\mathrm dx$ is also divergent. Your work is correct. Where did you see the claim that it was convergent? $\endgroup$
    – K.defaoite
    Sep 18, 2022 at 11:39

1 Answer 1


You are correct.

The $$\frac1{x^5}$$ part of your integral $$\int^1_{-1} \left(\frac 1 {x^5} + x^2\right)\ dx$$ is clearly divergent, as vertical asymptote we integrate through at the origin blows up too fast for any remote convergence. Only when the power is less than 1 would we have convergence, and clearly $5>1$. You can look at K.defaoite’s comment for a rigorous argument.

The solution on the other hand, seem to want to interpret this in a principal value sense.

Principal value is defined as follows. For a discontinuity at $b$ where $a<b<c$, a divergent integral can be interpreted as $$\lim_{ \; \varepsilon \to 0^+ \;} \, \left[ \, \int_a^{b-\varepsilon} f(x) \, \mathrm{d}x ~ + ~ \int_{b+\varepsilon}^c f(x) \, \mathrm{d}x \, \right]$$

With this, the principal value of the $$\frac1{x^5}$$ part here would indeed be $0$, and in this sense the integral is indeed convergent.

However, when we usually speak of divergences and convergencies of integrals, this is most definitely not factored in, and a solution that does would almost certainly and clearly state that it would be interpreting it in a principal value sense. Yours clearly does not.

Though it is to note that the solution seems to have kept it as two separate limits, which would give, as you have noted, an indeterminate form. If they were to do principal value correctly, it would have been a singular limit and it could be easily shown that the stuff inside the limit cancels and the limit goes to 0.

So well, the solution is basically wrong in both a regular sense and a principal value sense, so just ignore it. Your logic is correct.


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