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Write $y={\int }_C{e}^{xt}f(t)dt$ and move the factor $x$ onto $e^{xt}$ by integration by parts. The differential-equation residual isThus the Laplace integral solution is obtained fromup to a nonzero constant and branch choices. The endpoint factor is proportional to $t^b(1-t{)}^{a-b}$, so the given positive real parts ensure its vanishing at $0$ and $1$.

For $x>0$, choose the directed intervals $C_1=[0,1]$ and $C_2=(-\infty ,0]$; for $x<0$, choose $C_1=[0,1]$ and $C_2=[1,\infty )$. On each interval choose a continuous branch separately. Equivalently their integrands, after an irrelevant constant phase is removed, use $t^{b-1}(1-t{)}^{a-b-1}$ on $(0,1)$, $(-t{)}^{b-1}(1-t{)}^{a-b-1}$ on $t<0$, and $t^{b-1}(t-1{)}^{a-b-1}$ on $t>1$. Exponential decay at the infinite endpoint removes the remaining boundary term.

For large positive $x$, $C_1$ has a nonzero leading term proportional to $e^x{x}^{-(a-b)}$, while $C_2$ is proportional to $x^{-b}$. For large negative $x$, $C_1$ is proportional to $(-x{)}^{-b}$ and $C_2$ to $e^x(-x{)}^{-(a-b)}$. These distinct asymptotics prove independence in each case; the gamma factors are nonzero under the assumptions.

For $a=3,b=1$, integrate $e^{xt}(1-t)$ on $[0,1]$ to get $(e^x-1-x)/x^2$. A convenient independent pair on either side of zero isDirect substitution verifies both, and their ratio is not constant. Their singularities cancel in $y_2-y_1$. Since $e^x-1-x=x^2/2+O(x^3)$, the solution regular at zero with the required normalization isIts power series supplies the removable value and verifies the equation at the singular point.

For an admissible Laplace integral solution, differentiation under the integral sign and integration by parts giveChoose endpoints for which the boundary term vanishes. Setting the density in braces to zero givesTwo suitable contours are the rays ${\Gamma }_{+}:0\to +i\infty$ and ${\Gamma }_{-}:0\to -i\infty$, with a continuous square-root branch on each ray. The Gaussian decays at infinity, and $\zeta g=O({\zeta }^{1/2})$ tends to zero at the finite endpoint. The integrals converge for every finite $z$.

To check independence rather than merely exhibit different paths, defineWith unit normalization, the two solutions satisfy $w_{\pm }(0)=e^{\pm i\pi /4}{J}_0$ and $w_{\pm }^{\prime }(0)=\pm i{e}^{\pm i\pi /4}{J}_1$. Their Wronskian at zero is $-2i{J}_0{J}_1\ne 0$, so they form an independent pair.

Strictly, the printed necessity claim requires an injective contour-transform class as well as vanishing endpoint terms. An arbitrary contour representation may contain a null weight in a contour Laplace representation, so a nonzero solution alone does not force every representing $g$ to have the displayed form. The calculation constructs the canonical density and two valid independent solutions; it does not assume uniqueness for unrestricted contour densities.