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What test do I use to show that the following integral converges? If you could provide me with the process that leads to the answer that would really help.

  1. $\displaystyle \int_{0}^{1}\frac{x^n}{\sqrt{1-x^4}}\,dx$
  2. $\displaystyle \int_{0}^{\pi /2}\frac{\ln(\sin(x))}{\sqrt{x}}\,dx$


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My problem is to find a suitable function g (that converges) for the comparison test. – user78336 May 18 '13 at 23:35
up vote 1 down vote accepted

I like to use the comparison tests for these kind of problems.

For the first integral, use the fact that $x^n \leq x$ on $x \in [0,1]$ $\forall n \in \mathbb{N}$, and that $x^4<x^2$ on $[0,1]$, giving $$\frac{x^n}{\sqrt{1-x^4}} \leq \frac{x}{\sqrt{1-x^2}}$$ and use the fact that $f\leq g \implies \int f \leq \int g$ on any bounded interval.


$$\begin{align} \int_0^1 \frac{x^n}{\sqrt{1-x^4}} \mathrm{d}x &\leq \int_0^1 \frac{x}{\sqrt{1-x^2}} \mathrm{d}x \\ &=\left. -\sqrt{1-x^2} +c \right|_0^1 \\ &= 1 \end{align}$$

For the second use a similar idea, except here note that $\ln x < x$ $\forall x>0$ and that $\sin x< x$ for all $x$.

We do all this to get the integral of $g$ on the right side to be one which we can compute in closed form, which gives a bound on the integral of $f$ which you want to show converges, which is of course the standard comparison test.

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I agree it´s a good idea to use the comparison test. However I think that your function g doesn´t converges. We need to choose some g that converges to prove that f also converges. After that we can calculate lim (f/g). – user78336 May 18 '13 at 23:32
No, the integral of $g$ does converge:$$ \int_0^1 \frac{x}{\sqrt{1-x^2}} \mathrm{d}x = -\sqrt{1-x^2} +c |_0^1= 1$$ – Milind May 19 '13 at 3:37
OK then, my fault – user78336 May 19 '13 at 10:52

To check convergence at $x=0$, check that the integrand behaves as $x^a$, where $a>-1$. If there is something that looks like a singularity away from $x=0$, say at $x=x_0$, then substitute $y=x_0-x$. Further, $\ln{x}$ represents an integrable singularity.

For the first integral, you need to check the that singularity at $x=1$ is integrable; i.e., that, when you substitute $y=1-x$, then the integral behaves as $y^a$, where $a>-1$ at $y=0$. That is, check the behavior of


at $y=0$.

For the second integral, use the fact that $x^a \ln{x}$ is integrable at $x=0$ when $a>-1$. How does $\sin{x}$ behave near $x=0$?

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