- $y\geq0$ define $H(y)=\int_{z=1}^{\infty} \frac{1}{z^4+zy}\,dz$ Show that $H$ is a continuous function of $y$ and show $\lim\limits_{y \to +\infty}H(y)=0$.
Evaluate the integral $H(y)=\int_{z=1}^{\infty} \frac{1}{z^4+zy}\,dz$
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1nour: 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! – 2012-06-11
2 Answers
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 $\lim_{a \rightarrow \infty} f(a) = 0$.
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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1If DCT isn't your thing, just use AGM to see that $1/(x^4 + ax) \le 2/\sqrt{ax^5}$, so $f(a) \ll 1/\sqrt{a}$. – 2012-06-10