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For nonnegative integer $n$, a second solution of the Legendre differential equation is obtained by reduction of order from $P_n$ as $P_n(x)\int dx/[(1-x^2)P_n(x{)}^2]$. Normalize it as $Q_0=\frac{1}{2}\log |(1+x)/(1-x)|$ and $Q_1=x{Q}_0-1$, with the usual Legendre polynomial recurrence relation for higher degrees. Inside $(-1,1)$, $Q_n\sim -\frac{1}{2}\log (1-x)$ near $1$ and $Q_n\sim (-1{)}^n\frac{1}{2}\log (1+x)$ near $-1$. These follow from $P_n(1)=1$, $P_n(-1)=(-1{)}^n$ in the integral. Poles in the integral at interior polynomial zeros cancel in the product, by reduction of order across a zero of the known solution. On $x>1$, the integral from infinity defines the decaying normalization $Q_n(x)\sim 2^n(n!{)}^2{x}^{-n-1}/(2n+1)!$. For $n\ge 1$ all bounded exterior solutions decay this way; for $n=0$ an additional constant solution remains bounded.

The origin is an ordinary point, since the coefficients of the second-order linear differential equation are polynomials. Seek a power series $y={\sum }_{n\ge 0}{a}_n{x}^n$. Matching the coefficients of $x^0$ and $x^1$ gives $a_2=a_3=0$. Matching the coefficient of $x^{j+2}$ givesThus the four residue classes of indices modulo four evolve separately. The initial conditions for the first solution give $a_0=1$, $a_1=0$, and henceThe second set of initial conditions gives $a_0=0$, $a_1=1$, andAn empty product equals one. In either power series, the ratio of successive nonzero terms tends to zero for every fixed $x$: the coefficient ratio is of order $1/m$, while the power increases by four. The ratio test therefore proves convergence for every $x$, and termwise differentiation verifies the differential equation. These normalized solutions are linearly independent.

For their Wronskian $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$, differentiating and substituting the differential equation givesThus the Abel identity yieldsOn the interval containing zero where $y_1$ has no zero,Since $y_2(0)/y_1(0)=0$, integration proves the requested reduction of order formula:This is a local formula around the origin, as requested. The power series define both solutions globally; an ordinary integral through a zero of $y_1$ is not intended. Reduction of order across a zero of the known solution instead uses continuation of the actual solution.

Inspection gives $y_1=x$ in the degree-one Legendre differential equation. After division by $1-x^2$, $p=-2x/(1-x^2)$ and $e^{-\int p}=1/(1-x^2)$. On either side of zero, reduction of order therefore givesA second solution is $y_2=x\mathrm{arctanh}x-1$. Although $v$ is singular at zero, the product $v{y}_1$ has a smooth continuation throughout $(-1,1)$. This illustrates reduction of order across a zero of the known solution. Since $y_1(0)=0$, $y_1^{\prime }(0)=1$, $y_2(0)=-1$ and $y_2^{\prime }(0)=0$, the required coefficients are $1,-1$. HenceThe ordinary differential equation has regular coefficients near zero, so the resulting initial-value solution is unique and the removable singularity in the intermediate representation causes no ambiguity.

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 $Q_n$.