Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
  • $y\geq0$ define $$H(y)=\int_{z=1}^{\infty} \frac{1}{z^4+zy}\,dz$$ Show that $H$ is a continuous function of $y$ and show $\lim\limits_{y \to +\infty}H(y)=0$.
share|cite|improve this question
In general, if $f(x)$ is continuous in $[a,x]$, then $\int_a^x f(t)dt$ is continuous. Note that the integrand is continuous in $[1,+\infty]$. – Pedro Tamaroff Jun 10 '12 at 0:33
@PeterTamaroff Are you sure you read the problem correctly? $f(a)$ is not an antiderivative of $1/(x^4+ax)$. – Erick Wong Jun 10 '12 at 1:54
I'm puzzled about that as well. Are thinking in some change of variables? – leo Jun 10 '12 at 2:42
@Nour: To ask a new question, click on the ask question button. Do not edit existing questions to change them into new ones. I have rolled back your edit. – Eric Naslund Jun 10 '12 at 18:38
nour: You did it again! How are people supposed to know that $f$ is $H$ and $a$ is $y$ and that there is now a question about continuity? Please stop defacing your questions! – Did Jun 11 '12 at 5:28
up vote 1 down vote accepted

This is kind of a brute force method where we explicitly find the function $f(a)$. $$f(a) = \begin{cases} \dfrac{\log(1+a)}{3a} & \text{if }a >0\\ \dfrac13 & \text{if }a=0 \end{cases}$$ This can be obtained as shown below. We have that for $a>0$, $$\dfrac1{x^4+ax} = \dfrac1{x(x^3+a)} = \dfrac1{ax} - \dfrac{x^2}{a(a+x^3)}$$ Hence, $$f(a) = \int_1^{\infty} \dfrac{dx}{x^4+ax} = \int_1^{\infty} \left(\dfrac1{ax} - \dfrac{x^2}{a(a+x^3)} \right) dx = \lim_{R \rightarrow \infty} \int_1^{R} \left(\dfrac1{ax} - \dfrac{x^2}{a(a+x^3)} \right) dx$$ The first integral $$I_1 = \int_1^{R} \dfrac{dx}{ax} = \dfrac{\log(R)}a.$$ The second integral $$I_2 = \dfrac1{3a} \int_1^{R} \dfrac{3x^2dx}{(a+x^3)} = \left. \dfrac1{3a} \log(a+x^3) \right \rvert_{1}^{R} = \dfrac{\log(a+R^3) - \log(a+1)}{3a}$$ Putting these together, we get that \begin{align} f(a) & = I_1 - I_2\\ & = \lim_{R \rightarrow \infty} \left(\dfrac{\log(R)}a - \left( \dfrac{\log(a+R^3) - \log(a+1)}{3a}\right) \right)\\ & = \lim_{R \rightarrow \infty} \dfrac{\log(R^3)-\log(a+R^3) + \log(a+1)}{3a}\\ & = \lim_{R \rightarrow \infty} \dfrac{\log \left(\dfrac{R^3}{a+R^3} \right) + \log(a+1)}{3a}\\ & = \dfrac{\log (1) + \log(a+1)}{3a}\\ & = \dfrac{\log(a+1)}{3a}\\ \end{align} If $a=0$, then $f(0) = \displaystyle \int_1^{\infty} \dfrac{dx}{x^4} = \dfrac13$. Hence, we have that $$f(a) = \begin{cases} \dfrac{\log(1+a)}{3a} & \text{if }a >0\\ \dfrac13 & \text{if }a=0 \end{cases}$$ Clearly, $f$ is a continous function of $a$ for all $a \geq 0$ and $\lim_{a \rightarrow \infty} f(a) = 0$.

share|cite|improve this answer

You can apply here the rule for differentiating under the integral sign of Leibnitz:$$\frac{\partial f}{\partial a}=\int_1^\infty\frac{\partial}{\partial a}\left(\frac{1}{x^4+ax}\right)dx=\int_1^\infty\frac{-x}{(x^4+ax)^2}dx$$, and since $\,f(a)\,$ derivable then it is continuous.

For the limit you can use the dominated convergence theorem:$$a>0\,\,,\,x\in [1,\infty)\Longrightarrow\frac{1}{x^4+ax}\leq\frac{1}{x^4}\Longrightarrow $$$$\Longrightarrow\lim_{a\to\infty}\int_1^\infty\frac{1}{x^4+ax}dx=\int_1^\infty\lim_{a\to\infty}\frac{1}{x^4+ax}dx=0$$because $\,\displaystyle{\int_1^\infty\frac{1}{x^4}dx}\,$ exists

share|cite|improve this answer
Why we have to do DCT here? I think it is enough to write f(a) explicitly as done by Marvis and conclude that lim f(a)=0. Do you have a point to go to the DCT? Thank you – nour Jun 10 '12 at 12:40
Using the DCT is much easier. – Stefan Smith Jun 10 '12 at 13:04
@Nour Bless your heart if you think that what Marvis did is easier that using the DCT as above, though I might agree it is perhaps more elementary and, definitely, it ie enough. DCT was my first thought here and, as it happens, it can make things pretty easier. – DonAntonio Jun 10 '12 at 13:23
If DCT isn't your thing, just use AGM to see that $1/(x^4 + ax) \le 2/\sqrt{ax^5}$, so $f(a) \ll 1/\sqrt{a}$. – Erick Wong Jun 10 '12 at 19:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.