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

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