0
$\begingroup$

Would you please help me please solving this integral
$$ \int \exp( aZ^2 + bZ ) \, dZ $$

noting that $ a<0 $ , $b>0$

Thanks.

$\endgroup$
  • 3
    $\begingroup$ Complete the square to get an error function. $\endgroup$ – Jack D'Aurizio Nov 5 '15 at 19:38
  • $\begingroup$ As written I'm pretty sure the indefinite integral can't be done - perhaps you really meant the integral to have limits from $-\infty$ to $+\infty$ - that integral can be evaluated . $\endgroup$ – WW1 Nov 5 '15 at 19:45
  • $\begingroup$ Do you intend capital $Z$ and lower-case $z$ to refer to the same thing? If so, you shouldn't do that; if not, then your question is unclear. $\endgroup$ – Michael Hardy Nov 5 '15 at 19:55
0
$\begingroup$

When you see something like $aZ^2+bZ$ you immediately think that maybe completing the square is what you need: $$ aZ^2 + bZ = a\left( Z^2 + \frac b a Z \right) = a\left( Z^2 + \frac b a Z + \frac{b^2}{4a^2} \right) - \frac{b^2}{4a} = a\left( Z - \frac b {2a} \right)^2 - \frac{b^2}{4a} $$ Then \begin{align} & \int_{-\infty}^\infty \exp \left( aZ^2 + bZ \right) \, dZ = \int \exp \left( a\left( Z - \frac b {2a} \right)^2 - \frac{b^2}{4a} \right) \, dz \\[10pt] = {} & \exp\left( -\frac {b^2}{4a} \right) \int_{-\infty}^\infty \exp \left( a\left( Z - \frac b {2a} \right)^2 \right) \, dZ \end{align} Letting $W = Z - \dfrac b {2a}$ so that $dW = dZ$, we see that as $Z$ goes from $-\infty$ to $+\infty$, so does $W$, and we get $$ \exp\left( -\frac {b^2}{4a} \right) \int_{-\infty}^\infty \exp \left( aW^2 \right) \, dW. $$ Now letting $V= W\sqrt{-2a}$, so that $aW^2 = \frac{-1} 2 V^2$ and $dW = \dfrac{dV}{\sqrt{-2a}}$, we get $$ \frac {1} {\sqrt{-2a}} \exp\left( -\frac {b^2}{4a} \right) \int_{-\infty}^\infty \exp \left( -\frac 1 2 V^2 \right) \, dV. $$ Now you have the familiar Gaussian integral and you get

$$ \frac{1}{\sqrt{-2a}} \exp\left( -\frac {b^2}{4a} \right) \cdot \sqrt{2\pi}. $$

$\endgroup$
  • $\begingroup$ Thank you very much, I tried to use this result in order to solve the double integral which I am facing during my research..but I got an integral involving the Q-funciton.. Would you please help link $\endgroup$ – eMAS Nov 6 '15 at 8:21
1
$\begingroup$

Denote $$ \Phi(y) = \int_{-\infty}^y\exp\left(-\frac{1}{2}x^2\right) \mathrm{d}x $$ By construction $$ \Phi^\prime\left(y\right) = \exp\left(-\frac{1}{2} y^2 \right) $$ Now observe that, for $a>0$ $$ \frac{\mathrm{d}}{\mathrm{d} x} \Phi\left(\sqrt{2 a} x - b\right) = \sqrt{2 a} \Phi^\prime\left(\sqrt{2 a} x - b\right) = \sqrt{2 a} \exp\left(-\frac{1}{2} \left(\sqrt{2 a} x - b\right)^2 \right) $$ Now, expand the square: $$ \sqrt{2 a} \exp\left(-\frac{1}{2} \left(\sqrt{2 a} x - b\right)^2 \right) = \sqrt{2 a} \exp\left(-a x^2 + \sqrt{2 a} b x - \frac{1}{2} b^2 \right) $$ Hence $$ \exp\left(-a x^2 + c x\right) = \frac{\mathrm{d}}{\mathrm{d}x} \frac{\exp\left(\frac{c^2}{2a}\right)}{\sqrt{2a}} \Phi\left( \sqrt{2 a} x - \frac{c}{\sqrt{2a}} \right) $$ which is to say $$ \int \exp\left(-a Z^2 + c Z\right) \mathrm{d} Z = \frac{}{\sqrt{2a}} \exp\left(\frac{c^2}{2a}\right) \Phi\left( \sqrt{2 a} Z - \frac{c}{\sqrt{2a}} \right) + {\color\gray C} $$

$\endgroup$
0
$\begingroup$

Use partial integration on $e^{bz}$ and $\exp(az^2+bz)$ as follows: $$ \int{e^{az^2}e^{bz}dz}=\frac{e^{az^2}e^{bz}}{b}-\int{\frac{2ze^{az^2}e^{bz}}{b}dz}\\ \int{\frac{2ze^{az^2}e^{bz}}{b}dz}=\frac{2z}{b}\int{e^{az^2}e^{bz}dz}-\frac 2 b\iint{e^{az^2}e^{bz}dz^2} $$ Let $A=I'=B''=\exp(az^2+bz)$ so we have $$ \int Adz=\frac Ab-\left(\frac {2z}b\int Adz - \frac 2b\iint Adz^2\right)\\ \left(1+\frac 2bz\right)B' = \frac {B''}b + \frac {2B}b+C_0\\ B=\left(-\frac b2-z\right)B'+\left(-\frac 1 2\right)B'' +C_1\\ B'=\left(-\frac b2-z\right)B''+(-1)B'+\left(-\frac 1 2\right)B'''\\ I=\left(-\frac b4-\frac 1 2z\right)I'+\left(-\frac 1 4\right)I'' $$ Hopefully this is a solvable differential equation, maybe you know better than me?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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