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I have some trouble calculating this integral: $$\lim_{a\to 0} \int_0^2 {1 \over ax^4+2}\,\mathrm dx$$ Got something divided by zero all of the time.

Thanks in advance for any assistance!

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Why not push the limit inside the integral? –  hot_queen Jul 2 '13 at 11:48
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@hot_queen : You need a theorem for that and it doesn't work with all functions. –  xavierm02 Jul 2 '13 at 11:52
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@xavierno2: The theorem I used is called "Let us leave the OP to fill the details." –  hot_queen Jul 2 '13 at 11:58
    
In the range of integration, for $a>0$, ${1\over a2^4+2}\le {1\over ax^4+2}\le {1\over 2}$. Use the Squeeze Theorem to compute the limit from the right. Do something similar for the limit from the left. –  David Mitra Jul 2 '13 at 12:04
    
Hi, sadly we haven't shown/proven this theorem in class so I can't use it. –  ohad Jul 2 '13 at 13:17

3 Answers 3

We can do as hot_queen and Cocopuffs suggest, interchange limit and integration (using suitable justification, such as Lebesgue dominated convergence). This suggests that the result will be $1$.

We can also derive it a bit more elementarily. Note that:

$$\frac{1}{2+ax^4} = \frac12\left(1-\frac{ax^4}{2+ax^4}\right)$$

Hence:

\begin{align} \int_0^2 \frac{1}{2+ax^4}\,\mathrm dx = \int_0^2 \frac12\,\mathrm dx - \frac12 \int_0^2 \frac{ax^4}{2+ax^4}\,\mathrm dx \end{align}

and it will suffice to show that:

$$\lim_{a\to0}a \int_0^2 \frac{x^4}{2+ax^4}\,\mathrm dx = 0$$

which, due to the factor $a$, amounts to showing that:

$$\int_0^2\frac{x^4}{2+ax^4}\,\mathrm dx$$

is bounded in a suitable neighbourhood of $a = 0$. So assume that $|a| < 2^{-5}$; then for $0\le x \le 2$, $2+ax^4 \ge \frac32$, so that:

$$0 \le \frac{x^4}{2+ax^4} \le \frac23x^4$$

for $0 \le x \le 2$, which gives bounds on the integral:

$$0 \le \int_0^2 \frac{x^4}{2+ax^4}\,\mathrm dx \le \int_0^2 \frac23x^4\,\mathrm dx = \frac{64}{15}$$

Finally, we can conclude that:

$$0 \le \lim_{a\to0}a\int_0^2 \frac{x^4}{2+ax^4}\,\mathrm dx \le \lim_{a\to0}\frac{64}{15}a = 0$$

and for the original limit, that:

$$\lim_{a\to0}\int_0^2 \frac{1}{2+ax^4}\,\mathrm dx = \lim_{a\to0}\int_0^2 \frac12\,\mathrm dx - \lim_{a\to0}\frac12 \int_0^2 \frac{ax^4}{2+ax^4}\,\mathrm dx = 1 - 0 = 1$$

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You shouldn't be calculating any antiderivative of $(ax^4 + 2)^{-1}$. Instead, use the dominated convergence theorem and calculate $$\int_0^2 \lim_{a \rightarrow 0} \frac{1}{ax^4 + 2} \, \mathrm{d}x.$$

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I'm a bit hazy on the details*, but for convergence like this on a compact interval, a weaker theorem (assuming uniform convergence*) is enough... * the detail being, "justifying the uniform convergence" –  Clement C. Jul 2 '13 at 11:55
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@ClementC. This is true, uniform convergence on a compact interval is enough. It seemed easier to check the conditions for the DCT –  Cocopuffs Jul 2 '13 at 12:01

It is obvious that $$ I(a)=\int_0^2\frac{1}{ax^4+2}\,dx $$ converges for $a > -\frac18$ and diverges for $a \le -\frac18$.

Now, for every $a >-\frac18$ we have \begin{eqnarray} |I(a)-1|&=&|I(a)-I(0)|=\left|\int_0^2\left(\frac{1}{ax^4+2}-\frac12\right)\,dx\right|=\frac12\left|\int_0^2\frac{ax^4}{ax^4+2}\,dx\right|\\ &\le&\frac12\int_0^2\frac{|a|x^4}{|ax^4+2|}\,dx \le \frac{|a|}{4\min\{1,1+8a\}}\int_0^2x^4\,dx\\ &=&\frac{8|a|}{5\min\{1,1+8a\}}. \end{eqnarray} Since $$ \lim_{a \to 0}\frac{8|a|}{5\min\{1,1+8a\}}=0, $$ it follows that $$ \lim_{a\to 0} I(a)=I(0)=1. $$

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