Solving the limit of integrals $\lim\limits_{q \to 0}\int_0^1{1\over{qx^3+1}} \, \operatorname{d}\!x$

Question

how do I solve this one?

$$\lim_{q \to 0}\int_0^1{1\over{qx^3+1}} \, \operatorname{d}\!x$$

I tried substituting $t=qx^3+1$ which didn't work, and re-writing it as $1-{qx^3\over{qx^3+1}}$ and then substituting, but I didn't manage to get on.

Thanks in advance!

if you substitute $t=q^{\frac{1}{3}}x$ you get $\int \frac{dt}{t^3+1}$, which is integrable, but the result is really messy — Alex, Apr 19 '13 at 16:15

score 7 · Accepted Answer · answered Apr 19 '13 at 16:22

For any $q>0$, $$\frac{1}{q+1} = \int_0^1\frac{1}{q\cdot 1+1}\,\mathrm{d}x\leq\int_0^1\frac{1}{qx^3+1}\,\mathrm{d}x \leq \int_0^1\frac{1}{q\cdot 0+1}\,\mathrm{d}x = 1.$$ For any $q\in(-1,0)$, $$1 = \int_0^1\frac{1}{q\cdot 0+1}\,\mathrm{d}x\leq \int_0^1\frac{1}{qx^3+1}\,\mathrm{d}x\leq \int_0^1\frac{1}{q\cdot 1+1}\,\mathrm{d}x = \frac{1}{q+1}.$$

Thus, the limit for $q\to 0$ is $1$.

Wolphram jonny · Answer 2 · 2013-04-19T16:19:39.793

0

Try substituting $t=q^{1/3}x$. Or, you can put the limit inside the definite integral first (the easy way)

edited Apr 19 '13 at 16:19

answered Apr 19 '13 at 16:14

Wolphram jonny

1,950

Is that 'allowed'? – ohad Apr 19 '13 at 16:21
It is allowed, see http://en.wikipedia.org/wiki/Dominated_convergence_theorem, but I am not sure if you are supposed to assume you can (I mean, if it is homework). – Wolphram jonny Apr 19 '13 at 16:23
Sadly we didn't prove this theorom in class so I still can't use it. Thanks anyway! – ohad Apr 19 '13 at 16:31

Solving the limit of integrals $\lim\limits_{q \to 0}\int_0^1{1\over{qx^3+1}} \, \operatorname{d}\!x$

2 Answers2