The first term in Watson's expansion amounts to the saddlepoint approximation (expansion of the exponent to second order around $t=1$), which gives $$I_{sp}=\frac{1}{2}e^{-x}\int_{-\infty}^\infty e^{-(x/2)(1-t)^2}\,dt=\frac{\sqrt{\pi } e^{-x}}{\sqrt{2 x}}.$$ This is quite accurate, see the plot (gold: exact; blue: saddlepoint approximation)
The saddlepoint approximation (expansion of the exponent to second order around $t=1$) gives $$I_{sp}=\frac{1}{2}e^{-x}\int_{-\infty}^\infty e^{-(x/2)(1-t)^2}\,dt=\frac{\sqrt{\pi } e^{-x}}{\sqrt{2 x}}.$$ This is quite accurate, see the plot (gold: exact; blue: saddlepoint approximation)
The first term in Watson's expansion amounts to the saddlepoint approximation (expansion of the exponent to second order around $t=1$), which gives $$I_{sp}=\frac{1}{2}e^{-x}\int_{-\infty}^\infty e^{-(x/2)(1-t)^2}\,dt=\frac{\sqrt{\pi } e^{-x}}{\sqrt{2 x}}.$$ This is quite accurate, see the plot (gold: exact; blue: saddlepoint approximation)
The saddlepoint approximation (expansion of the exponent to second order around $t=1$) gives $$I_{sp}=\frac{1}{2}e^{-x}\int_{-\infty}^\infty e^{-(x/2)(1-t)^2}\,dt=\frac{\sqrt{\pi } e^{-x}}{\sqrt{2 x}}.$$ This is quite accurate, see the plot (gold: exact; blue: saddlepoint approximation)
