Generals 2002 II 3

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The equation is:

I\left(x\right)=\int_{0}^{1}\left(1+t\right)e^{x\sin\left(3t\right)}dt


If we write \phi\left(x,t\right)=x\sin\left(3t\right)+\ln\left(1+t\right), then we can take the derivative to find the saddles:

\phi^{\prime}=3x\cos3t+\frac{1}{1+t}=0


Which can be simplified for large x and t from 0 to 1 to be approximately:

cos3t = 0


Then the only saddle is at t = π / 6. The second derivitive is:

\phi_{0}^{\prime\prime}\approx-9x\sin\frac{\pi}{2}=-6x


So the direction is \sqrt{-\phi_{0}^{\prime\prime}}\sim\sqrt{9x}, so it is real (for positive real x), and we can integrate through it. Doing this, we get:

I\left(x\right)\approx\frac{\sqrt{2\pi}}{\sqrt{9x}}\left(1+\frac{\pi}{6}\right)e^{x}
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