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For the Legendre differential equation $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+n(n+1)y=0$, put $t=1+x$. The equation becomes $t(2-t)y_{tt}+2(1-t)y_t+n(n+1)y=0$. Its indicial equation is $2r^2=0$. The exponent-zero power series $\sum a_k{t}^k$ obeys the displayed linear recurrence relation. For a nonnegative integer $n$, it terminates at degree $n$, giving a Legendre polynomial with $a_0\ne 0$. Every analytic solution at the endpoint is proportional to this one, since the recurrence determines all coefficients from $a_0$.

The second independent solution has the local form $P_n(x)\log (1+x)+B(x)$, where $B$ is analytic near $-1$. Indeed, reduction of order gives $P_n(x)\int dx/[(1-x^2)P_n(x{)}^2]$, whose integrand is $1/[2P_n(-1{)}^2(1+x)]$ plus an analytic function. Rescale the second solution to obtain the unit coefficient of $P_n\log (1+x)$. Reflection gives the corresponding expansion at $x=1$. This also proves the logarithmic divergence of the independent Legendre function of the second kind at each endpoint.

Use $t=x+1$. The Legendre differential equation becomesSubstituting a Frobenius series $y=t^r{\sum }_{k\ge 0}{a}_k{t}^k$, $a_0\ne 0$, gives the indicial equation $2r^2=0$: the indicial root zero is repeated. For $r=0$, the coefficient of $t^k$ yieldsorFor $0\le k<n$ the multiplying factor is nonzero. At $k=n$ it vanishes, so $a_{n+1}$ and all later coefficients are zero. Thus one solution is a polynomial of exact polynomial degree $n$, including a nonzero constant when $n=0$. Denote a chosen normalization by $P_n$.

The linear recurrence relation determines every analytic coefficient from $a_0$, so there cannot be a second independent analytic solution at $t=0$. To see the required logarithm explicitly, reduction of order gives the second solution locally asNear $t=0$, the integrand is $1/(2a_0^{2}t)$ plus an analytic function. After rescaling the second solution, its form is thereforeThe analytic remainder converges locally; its constant term may be set to zero by adding a multiple of $P_n$. The nonzero logarithmic coefficient makes $y_2$ linearly independent of $P_n$. This Frobenius expansion of the Legendre equation at an endpoint gives a Logarithmic Frobenius solution; the same endpoint calculation at $x=1$ explains why the independent Legendre function of the second kind is excluded when boundedness at both ends is required.

When $n=0$, the forcing has nonzero integral ${\int }_{-1}^{1}1dx=2$, violating the Fredholm alternative condition. There is no bounded solution. Directly, integration gives $(1-x^2)y^{\prime }=x+B$; cancellation of its logarithmic divergence at $x=-1$ requires $B=1$, while cancellation at $x=1$ requires $B=-1$.

For $n\ge 1$, the constant $1/[n(n+1)]$ is a particular solution. Subtract it from any solution. The remainder satisfies the Legendre differential equation, whose general solution is a combination of $P_n$ and the logarithmically unbounded Legendre function of the second kind. Boundedness removes the latter coefficient, givingThe constant forcing is orthogonal to $P_n$ for $n\ge 1$, consistently with Orthogonality of Legendre polynomials and the Fredholm alternative.

For a nonnegative integer $\ell$, the chain of the same parity as $\ell$ reaches $n=\ell$, where the numerator of the recurrence relation vanishes. The next coefficient and every later coefficient in that chain are zero, producing a polynomial of degree $\ell$. Normalizing it at $x=1$ gives the Legendre polynomial $P_{\ell }$.

The opposite-parity chain never reaches $n=\ell$. Its other possible zero, $n=-\ell -1$, is negative and cannot occur, so a nonzero initial coefficient produces a nonterminating power series. It is linearly independent of $P_{\ell }$ and can be scaled to the Legendre function of the second kind $Q_{\ell }$.

The degree statement requires $\ell \ge 0$, as the examples in part (c) indicate. For a negative integer $\ell$, the same equation has the nonnegative degree $-\ell -1$, because $\ell (\ell +1)=(-\ell -1)(-\ell )$. A polynomial cannot literally have negative degree.

Since $P_0=1$, part (d) gives $Q_0^{\prime }=A_0/(1-x^2)$. A partial fraction decomposition and integration give, for $-1<x<1$,The condition at zero fixes $B=0$. Thus the full conclusion from the stated condition isThat condition alone does not fix the multiplicative constant. With the usual Legendre function of the second kind normalization $Q_0^{\prime }(0)=1$, one has $A_0=1$. On real exterior intervals replace the logarithm by $\log |(1+x)/(1-x)|$; complex continuation requires a choice of logarithm branch.

Two integrations give the general reduction of order formulaThe integral is local on intervals avoiding zeros of $P_n$. A pole of the integral at an interior polynomial zero need not be a singularity of $y$: the product extends by reduction of order across a zero of the known solution. After a standard normalization the independent second solution is the Legendre function of the second kind $Q_n$.

For normalized Legendre polynomials, $P_n(1)=1$ and $P_n(-1)=(-1{)}^n$. Near the endpoints the integral factor has the asymptoticsandThus an independent solution, for which $C_2\ne 0$, has the logarithmic endpoint singularitiesThe constants are understood for a solution continued consistently across the ordinary interior points. Adding any multiple of $P_n$ cannot cancel these logarithms. This explains why boundedness at an endpoint selects the Legendre polynomial and excludes the independent Legendre function of the second kind.