Definite Integral over Reals of Exponential of -(a x^2 plus b x plus c)
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Theorem
- $\ds \int_{-\infty}^\infty \map \exp {-\paren {a x^2 + b x + c} } \rd x = \sqrt {\frac \pi a} \map \exp {\frac {b^2 - 4 a c} {4 a} }$
where $a$, $b$ and $c$ are real numbers with $a > 0$.
Proof
\(\ds \int_{-\infty}^\infty \map \exp {-\paren {a x^2 + b x + c} } \rd x\) | \(=\) | \(\ds \int_{-\infty}^\infty \map \exp {-a \paren {x + \frac b {2 a} }^2 + \frac {b^2} {4 a} - c} \rd x\) | Completing the Square | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \exp {\frac {b^2 - 4 a c} {4 a} } \int_{-\infty}^\infty \map \exp {-a \paren {x + \frac b {2 a} }^2} \rd x\) | Exponential of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \exp {\frac {b^2 - 4 a c} {4 a} } \int_{-\infty}^\infty \map \exp {-\paren {\sqrt a x + \frac b {2 \sqrt a} }^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt a} \map \exp {\frac {b^2 - 4 a c} {4 a} } \int_{-\infty}^\infty \map \exp {-u^2} \rd u\) | substituting $u = \sqrt a + \dfrac b {2 \sqrt a}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\frac \pi a} \map \exp {\frac {b^2 - 4 a c} {4 a} }\) | Gaussian Integral |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Exponential Functions: $15.75$