# Evaluate the integral $H(y)=\int_{z=1}^{\infty} \frac{1}{z^4+zy}\,dz$

• $y\geq0$ define $$H(y)=\int_{z=1}^{\infty} \frac{1}{z^4+zy}\,dz$$ Show that $H$ is a continuous function of $y$ and show $\lim\limits_{y \to +\infty}H(y)=0$.
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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

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 $\lim_{a \rightarrow \infty} f(a) = 0$.

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$\newcommand{\d}{\mathrm d}$ Notice that $$\frac{\d}{\d a}\frac{1}{x^4+ax}=-\frac{x}{(x^4+ax)^2}$$ so, if we define $g(a)=\frac{1}{x^4+ax}$ for any $a\geq 0$, whenever $x\in[1,\infty[$ we have $$g'(a)\leq \left| \frac{1}{x^7} \right|,$$ and by the mean value theorem this yields $$|g(a_1)-g(a_2)|\leq |g'(\theta)||a_1-a_2|\leq \frac{|a_1-a_2|}{x^7}\quad\text{ for some }\theta\text{ between } a_1\text{ and }a_2.$$ Therefore $$|f(a_1)-f(a_2)|\leq|a_1-a_2|\int_1^\infty \frac{\d x}{x^7}=|a_1-a_2|,$$ so $f$ is Lipschitz and then continuous.

Concerning to the limit, it's enough to show that for any sequence $\{a_n\}_{n\in\Bbb N}$ of nonegative numbers, $\lim_{n\to\infty}a_n=\infty$ implies $f(a_n)\to 0$.

So, consider $\{a_n\}_{n\in\Bbb N}$ such a sequence. Since the $a_n$ are nonnegative, for $x\geq 1$, the inequalities $$x^4+a_nx\geq a_nx\geq a_n\geq 0$$ gives $$\frac{1}{x^4+a_nx}\leq \frac1{a_nx}\leq\frac1{a_n}$$ for $n$ large enough. This says that the sequence of functions $\left\{\frac{1}{x^4+a_nx}\right\}_{n\in\Bbb N}$ converges uniformly to $0$ in $[1,\infty[$. So you can move the limit inside the integral to get $$\lim_{n\to\infty} f(a_n)=0$$ as we wanted.

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Do you really think that if $f_n(x)\to0$ uniformly on $x\geqslant1$ then $\int\limits_1^{+\infty}f_n(x)\mathrm dx\to0$? –  Did Jun 10 '12 at 7:58
@did you're right. Let me see what can I do. –  leo Jun 10 '12 at 16:28

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

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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