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This integral I got from an MIT Bee Integral contest. The actual question is:

$$\displaystyle \int_{-\infty}^{\infty} e^{-2x^2-5x-3} \mathrm dx$$

So I look at the exponent and I see the factors so why not?

$\displaystyle \int e^{-(x+1)(2x+3)} \mathrm dx$

Then I go for the substitution:

$\begin{align}u&=x+1 \\ \mathrm du&=\mathrm dx \end{align}$

So we turn this into:

$\displaystyle \int e^{-u(2u+1)} \mathrm du$

$\displaystyle \int e^{-u} \cdot e^{-2u^2} \mathrm du$

So far so good. Yet another substitution because of some potential I saw.

$\begin{align} v&=e^{-u} \\ u&=-\ln v \\ \mathrm du&= -\dfrac{1}{v} \mathrm dv \end{align}$

The next transformation is:

$\displaystyle \int v \cdot e^{-2\ln^2 v} \cdot -\dfrac{1}{v} \mathrm dv$

$=\displaystyle \int -e^{-2\ln^2 v} \mathrm dv$

This is where I saw blanks. My question is: is this even the right way to go about this? If so please help out from here, otherwise any recommendations or ideas would be appreciated.

Lastly, if the method is easy for someone with scraps of Calc 2 knowledge just leave a hint because I want to own this thing. Thanks :)

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    $\begingroup$ It's not, the best thing to do would be to complete the square then pull out the constants and translate the resulting Gaussian. $\endgroup$ – Ninad Munshi Apr 11 '20 at 7:48
  • $\begingroup$ @ Ninad Munshi From the beginning or some point in the middle. $\endgroup$ – Nεo Pλατo Apr 11 '20 at 7:49
  • $\begingroup$ @Jyrki Lahtonen Yes it does. Thanks $\endgroup$ – Nεo Pλατo Apr 11 '20 at 8:03
  • $\begingroup$ I don't really see the point of integration bees in particular, but I would think this is an integral you have to know to do well in such contests. Good luck. $\endgroup$ – Jyrki Lahtonen Apr 11 '20 at 8:07
  • $\begingroup$ This practice is more for fun than preparation for an actual contest. Thank you nevertheless $\endgroup$ – Nεo Pλατo Apr 11 '20 at 8:09
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$$-(2x^2+5x+3) = -2\left(x+\frac54\right)^2 +\frac18 $$

So,

$$\int_{-\infty}^{\infty}e^{-(2x^2+5x+3)}dx = e^{\frac18}\int_{-\infty}^{\infty}e^{-2\left(x+\frac54\right)^2}dx = e^\frac18\sqrt{\frac\pi2}$$

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Complete the square in the exponent instead to get the integral

$$\int_{-\infty}^\infty e^{-2\left(x+\frac{5}{4}\right)^2+\frac{1}{8}}\:dx = e^{\frac{1}{8}}\int_{-\infty}^\infty e^{-2t^2}\:dt = e^{\frac{1}{8}}\sqrt{\frac{\pi}{2}}$$

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  • $\begingroup$ What I'm guessing from your answer is that $\displaystyle \int_{-\infty}^{\infty}e^{-2x^2} dx$ has a known value. Is that right? $\endgroup$ – Nεo Pλατo Apr 11 '20 at 7:55
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    $\begingroup$ @Plato yes, the integral has a known value of $$\int_{-\infty}^\infty e^{-ax^2}\:dx = \sqrt{\frac{\pi}{a}}$$ $\endgroup$ – Ninad Munshi Apr 11 '20 at 7:56
  • $\begingroup$ @ Ninad Munshi In that case I am not at fault. Let me try some others and see what happens. $\endgroup$ – Nεo Pλατo Apr 11 '20 at 7:57
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    $\begingroup$ @Plato Good luck! If you would like to try some problems I wrote for UC Berkeley's first integration bee this past January, I would be more than happy to send them. $\endgroup$ – Ninad Munshi Apr 11 '20 at 7:58
  • $\begingroup$ @Plato, also, if you have to accept an answer, choose Ak19's since they got the constant right the first time. $\endgroup$ – Ninad Munshi Apr 11 '20 at 7:59

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