Explicit functions evaluated (a) Defined $f$ by $f(y):=\int_0^\infty\frac{xy}{(x^4+y^4)^{3/4}}dx$. Prove $f(y)$ is defined (i.e integral exists) for every $y\in\mathbb{R}$.
(b)Prove that actually $f(y)=c\operatorname{sign} y$ for some positive number $c$. *So in particular $f$ is not continuous at $0$, therefore a substitution may help and no need to evaluate $c$ explicitly.
(c)Prove that $g(y):=\int_0^\infty\frac{xy}{(x^4+y^4+x^2)^{3/4}}dx$ DOES define a function that is continuous for all $y\in \mathbb{R}$.
These are a few examples I ran across in the midst of a self study. Please, any help is appreciated.

Part (a)
Clearly $f(0) = 0$.
For $x \geqslant 1,$
$$\left|\frac{xy}{(x^4+y^4)^{3/4}} \right| = \frac{x|y|}{(x^4+y^4)^{3/4} } \leqslant \frac{x|y|}{x^3 } = \frac{|y|}{x^2},$$
and, since,
$$\int_1^\infty \frac{dx}{x^2} = 1 < \infty,$$
it follows by the comparison test that 
$$\int_1^\infty\frac{xy}{(x^4+y^4)^{3/4}} \, dx < \infty$$
For $0 < x \leqslant 1$ and $y \neq 0,$
$$\left|\frac{xy}{(x^4+y^4)^{3/4}} \right| = \frac{x|y|}{(x^4+y^4)^{3/4} } \leqslant \frac{x|y|}{|y|^3 } = \frac{x}{|y|^2},$$
and, since, $x$ is integrable over $[0,1],$ it follows, again, by the comparison test that 
$$\int_0^1\frac{xy}{(x^4+y^4)^{3/4}} \, dx < \infty$$
Therefore, $f(y)$ is defined for all $y \in \mathbb{R}.$
Part (b)
We have
$$f(y) = \int_0^\infty \frac{xy}{(x^4+y^4)^{3/4}}  \, dx = \int_0^\infty \frac{xy}{|y|^3\left(1+ x^4/|y|^4\right)^{3/4}}  \, dx. $$
Changing variables with $x = |y|u$ we have
$$f(y) = \frac{y}{|y|^3} \int_0^\infty \frac{|y|u}{\left(1+ u^4\right)^{3/4}}|y|  \, du \\ = \frac{y}{|y|} \int_0^\infty \frac{u}{\left(1+ u^4\right)^{3/4}}  \, du \\ = \frac{y}{|y|} \int_0^\infty \frac{u}{\left(1+ u^4\right)^{3/4}}  \, du \\ = c\text{sign}(y),$$
where
$$c = \int_0^\infty \frac{u}{\left(1+ u^4\right)^{3/4}}  \, du.$$
Therefore, $f$ is not continuous at $y =0$ since $f(0) = 0$ and
$$ \lim_{ y \to 0+}f(y) = c, \\  \lim_{ y \to 0-}f(y) = -c.$$
Part (c)
Again we have $g(0) = 0.$ For $y$ in any compact interval $[-b,b],$ we find
$$\left|\frac{xy}{(x^4+y^4+x^2)^{3/4}}\right| \leqslant h(x) = \begin{cases} \frac{|b|}{x^2}, \,\,\,x \geqslant 1 \\ \frac{|b|}{\sqrt{x}}, \,\, 0 < x \leqslant 1\end{cases}.$$
Since $h$ is integrable over $[0, \infty)$ it follows by the Lebesgue dominated convergence theorem that $g$ is continuous for all $y \in \mathbb{R},$ since for any $y_0 \in [-b,b]$ where $b$ can be as large as we like, 
$$\lim_{y \to y_0} g(y) = \int_0^\infty \lim_{y \to y_0} \frac{xy}{(x^4+y^4+x^2)^{3/4}}\, dx  = g(y_0).$$
