Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I integrate,

$$ \int_n^{+\infty} x \exp\{-ax^2+bx+c\}dx $$

and what's the result w.r.t the Gaussian function's p.d.f $p(x)$ and c.d.f $\phi(x)$?


share|cite|improve this question
@DilipSarwate c.d.f of course. Thanks for pointing out. Edited. – shuaiyuancn Apr 17 '12 at 20:56
up vote 2 down vote accepted

Completing the square in the quadratic: $$\eqalign{ \int_n^\infty x \exp(-ax^2+bx+c)\,dx &= \int_n^\infty\kern-5pt x \exp( -a(x-{\textstyle{b\over 2a})^2 +c+{b^2\over 4a} } )\,dx\cr &= \alpha \int_n^\infty\kern-5pt x \exp( \textstyle {-(x-{\textstyle{b\over 2a})^2 }\over 1/a } )\,dx\cr &= \alpha \int_n^\infty\kern-5pt (x+{\textstyle{b\over 2a}-{b\over2a}}) \exp( \textstyle {-(x-{\textstyle{b\over 2a})^2 }\over 1/a } )\,dx\cr &=\alpha \int_n^\infty\kern-5pt (x { -{\textstyle{b\over2a}}}) \exp( {\textstyle {-(x-{\textstyle{b\over 2a})^2 }\over 1/a }} )\,dx + \alpha\int_n^\infty\kern-5pt \textstyle{b\over2a} \exp( \textstyle {-(x-{\textstyle{b\over 2a})^2 }\over 1/a } )\,dx,\cr } $$ where $\alpha=\exp(c+{b^2\over4a})$.

On the right hand side of the last equality above, the first integral can be evaluated using the substitution $u=x-{b\over 2a}$ and the second integral can be expressed in terms of the cumulative distribution function of an appropriate normal random variable.

share|cite|improve this answer

Hint: Since $\frac{\mathrm d}{\mathrm dx}\exp(-ax^2+bx+c) = (-2ax+b)\exp(-ax^2+bx+c)$, you can massage the given integrand to something of the form $$\frac{-1}{2a}\int (-2ax+b)\exp(-ax^2+bx+c)\ \mathrm dx + \int \frac{b}{2a}\exp(-ax^2+bx+c)\ \mathrm dx$$ where the first integral now has a perfect integrand and the second, after further massaging will give you something involving $\Phi(\cdot)$, the cdf of the standard normal (Gaussian) random variable. Note that @Wonder's answer does not get the second term.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.