Integral equations and the Fredholm alternative / theory The Fredholm alternative states that either:
$$ 0 = \lambda \phi(x) - \int_a^b K(x,y) \phi(y) dy $$
has a non-trivial solution, or:
$$ f(x) = \lambda \phi(x) - \int_a^b K(x,y) \phi(y) dy $$
always has a unique solution for any f(x)
A sufficient condition is for the kernel K to be square-integrable, but depending on sources there is some confusion whether $\lambda$ must be a non-zero complex number.  For example, Wikipedia says so but also refers to this page which doesn't...
(Note that $\lambda=0$ the integral equation is called Fredholm of the first kind, and $\lambda \neq 0$ is of second kind.)
Adding to the confusion, some authors prefer to have $\lambda$ scaling the integral rather than $\phi(x)$, in which case it makes sense to require it to be nonzero.
Could anyone clarify whether the Fredholm alternative is only true of second kind equations (i.e. $\lambda \neq 0$)?  And if so, why...?
Thanks!
 A: In abstract terms, what is important is that the operator $\mathcal K\colon L^2(a, b)\to L^2(a, b)$ defined by 
$$\mathcal K\phi(x)=\int_a^b K(x, y)\phi(y)\, dy,$$
is compact. Such operators are never surjective. Therefore, for the Fredholm equation of the first kind 
$$
\mathcal K \phi = f $$ 
there always exists some $f\in L^2$ such that the equation has no solution.
CONCRETE EXAMPLE. Let 
$$
K(x, y)=\begin{cases} 
1, & x>y, \\ 0, & x\le y, \end{cases}$$
so that $\newcommand{\Kcal}{\mathcal{K}} \Kcal\phi(x)=\int_{-1}^x\phi(y)\, dy$. Here $(a, b)=(-1,1)$. Notice that, if $\phi\in L^2(-1,1)$, then $\Kcal\phi$ is a continuous function (actually, $\phi\in L^1(-1,1)$ would suffice). Therefore, if $f$ is not continuous, then the equation $\Kcal \phi =f$ has no solutions. 
Let us now add a tiny $\lambda \ne 0$ and consider the equation 
$$
\Kcal \phi + \lambda \phi = f.$$
The Fredholm theory guarantees that this equation has a unique solution for all $f\in L^2(-1, 1)$. Let us plug in a discontinuous function, such as 
$$
f(x)=H(x)=\begin{cases} 1, & x>0, \\ 0, & x\le 0,\end{cases}$$ 
($H$ stands for Heaviside), and see what happens. Taking a formal derivative, the equation becomes 
$$
\phi + \lambda \phi' = \delta(x), $$ 
where $\delta$ is the Dirac distribution. Now, $$\phi(x)=e^{-x/\lambda} H(x)$$ does satisfy this equation, because, if we plug it into the left-hand side, we get
$$
e^{-\frac x \lambda} H(x) -\frac{\lambda}{\lambda} e^{-\frac x \lambda} H(x) +e^{-\frac x \lambda} \delta(x)= \delta(x), $$ 
where we used the identity $ e^{-\frac x \lambda} \delta(x)= e^{-0}\delta(x)=\delta(x)$. 
(WARNING! This is not 100% rigorous, as we should now verify that $\phi$ solves the integral equation).
Summarizing, the presence of that parameter $\lambda\ne 0$, no matter how tiny it is, enabled the surjectivity of the operator. If the parameter tends to $0$, however, the solution $\phi$ we just found ceases to have a meaning, compatibly with the fact that $\Kcal$ cannot be surjective.
A REFERENCE. You asked for a reference. The book of Eidelman, Milman and Tsolomitis is one, chapter 5. See pag.82 for a remark on the parameters $\lambda$ and $\mu$, and the relationship between the classical notation on integral equation and the more abstract notation with the compact operators.
