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Gaussian integral 2016, Part II

The Gaussian integral states that +ex2dx=π \displaystyle \int_{-\infty}^{+\infty} e^{-x^2 } \, dx = \sqrt{\pi } .

Hence or otherwise, evaluate the integral

+ex2+2x+2016dx\large \int_{-\infty}^{+\infty} e^{-x^2 + 2x + 2016} \, dx

If the above integral can be expressed in the form aebπ a e^{b} \sqrt{\pi} , where a,ba , b are positive integers, find a+b a+b.


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