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Is there an easy way to compute $$\int_{-\infty}^\infty\exp(-x^2+2x)\mathrm{d}x$$ without using a computer package?

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    $\begingroup$ Try completing the square in the exponent. $\endgroup$ – Thomas Belulovich Jun 28 '12 at 10:34
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    $\begingroup$ $\exp(-x^2+2x)=\exp(-(x-1)^2)/e$, and substituting with $u=x-1$. $\endgroup$ – Yai0Phah Jun 28 '12 at 10:35
  • $\begingroup$ Thank you both for your comments, I've just figured it out. $\endgroup$ – James Jun 28 '12 at 10:37
  • $\begingroup$ I judt ran into this. Found a solution in larsons problem solving through problems. Problem 1 ,4 ,4. He squares entire integral. google.com/url?sa=t&source=web&rct=j&url=https://… $\endgroup$ – Marcus Watt Jul 21 '18 at 6:51
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In general $$ \begin{align} \int_{x=0}^\infty e^{-(ax^2+bx)}\,dx&=\int_{x=0}^\infty \exp\left(-a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\ \end{align} $$ Let $u=x+\frac{b}{2a}\;\rightarrow\;du=dx$, then $$ \begin{align} \int_{x=0}^\infty e^{-(ax^2+bx)}\,dx&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\ &=\exp\left(\frac{b^2}{4a}\right)\int_{u=0}^\infty e^{-au^2}\,du.\\ \end{align} $$ The last form integral is Gaussian integral that equals to $\frac{1}{2}\sqrt{\frac{\pi}{a}}$. Hence $$ \int_{x=0}^\infty e^{-(ax^2+bx)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right). $$ In your case, $$ \begin{align} \int_{-\infty}^\infty e^{-x^2+2x}\,dx&=2\int_0^\infty e^{-(x^2-2x)}\,dx\\ &=2\cdot\frac{1}{2}\sqrt{\frac{\pi}{1}}\exp\left(\frac{(-2)^2}{4\cdot1}\right)\\ &=e\sqrt{\pi}. \end{align} $$


$$\text{# }\mathbb{Q.E.D.}\text{ #}$$

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  • $\begingroup$ I don't understand why the lower limit $x = 0$ does not change to $u = \frac{b}{2a}$ up on the change of variables. Indeed wolfram alpha shows your answer is not correct, likely due to this limit change? $\endgroup$ – Señor O Nov 21 '17 at 4:48
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This is a Gaussian integral. In general you can use the formula $\int_{-\infty}^{\infty} \exp(-x^2+bx+c)\mathrm{d}x=\sqrt{\pi}~\exp(b^2/4+c)$. This formula, as suggested by Thomas, can be derived by completing the square in the exponent, and using $\int_{-\infty}^{\infty} \exp(-x^2)\mathrm{d}x=\sqrt{\pi}$.

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