Gaussian integral 2016, Part I

Calculus Level 3

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

Given that information (if it is related at all to this problem), evaluate the integral

∫ − ∞ + ∞ e − 25 9 ( x + 2016 ) 2 d x \displaystyle \int_{- \infty}^{+\infty} e^{-\frac{25}{9} (x+2016)^{2}} \, dx

If the integral above can be expressed in the form

a b π \dfrac{a}{b} \sqrt{\pi }

with a , b a, b as positive coprime integers, find a + b a+b .


The answer is 8.

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Hobart Pao by Hobart Pao , 23, USA

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1 solution

Sam Bealing
Apr 3, 2016

As the limits are to infinity, x + 2016 x+2016 doesn't affect the integral so we can say:

∫ − ∞ ∞ e − 25 9 ( x + 2016 ) 2 d x = ∫ − ∞ ∞ e − 25 9 ( x ) 2 d x = ∫ − ∞ ∞ e − ( 5 x 3 ) 2 d x \int_{-\infty }^{\infty} e^{-\frac{25}{9}(x+2016)^2} dx= \int_{-\infty }^{\infty} e^{-\frac{25}{9}(x)^2} dx=\int_{-\infty }^{\infty} e^{-(\frac{5x}{3})^2} dx

Substituting u = 5 x 3 ⇒ 3 5 d u = d x u=\frac{5x}{3} \Rightarrow \frac{3}{5} du=dx gives:

∫ − ∞ ∞ e − ( 5 x 3 ) 2 d x = 3 5 ∫ − ∞ ∞ e − u 2 d u = 3 5 π \int_{-\infty }^{\infty} e^{-(\frac{5x}{3})^2} dx=\frac{3}{5}\int_{-\infty }^{\infty} e^{-u^2} du=\frac{3}{5}\sqrt{\pi}

So a = 3 , b = 5 ⇒ a + b = 8 a=3,b=5 \Rightarrow a+b=8

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