1
$\begingroup$

I have been able to solve:

$\int_0^{+\infty} ue^{-Au^2+Bu}du, ~A>0, B>0$ by means of symbolic calculus library such as Mathematica or Sympy.

However, I am wondering what is the classical method to solve such a definite integral.

I have also looked in the Gradshteyn and Ryzhik's Table of Integrals, Series, and Products, but I did not find it as only the result of $\int_{-\infty}^{+\infty} ue^{-Au^2+Bu}du$ is presented.

$\endgroup$
1
  • $\begingroup$ i think this is a solution $$\frac{\sqrt{\pi } b e^{\frac{b^2}{4 a}} \left(\text{erf}\left(\frac{b}{2 \sqrt{a}}\right)+1\right)+2 \sqrt{a}}{4 a^{3/2}}$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 7 '17 at 11:59
1
$\begingroup$

Start by completing the square in the exponent:

$$-Au^2+Bu=u(-Au+B)=-A(u-B/2A)^2+B^2/4A$$

and substituting $x=u-a$ where $a=-B/2A$ to get

$$I=e^{B^2/4A}\int_a^\infty(x-a)e^{-Ax^2}~\mathrm dx$$

Note that

$$\int_a^\infty xe^{-Ax^2}~\mathrm dx=\frac12\int_{a^2}^\infty e^{-At}~\mathrm dt=\frac1{2A}e^{-Aa^2}$$

and

$$\int_a^\infty e^{-Ax^2}~\mathrm dx=\frac1{\sqrt A}\int_{a\sqrt A}^\infty e^{-t^2}~\mathrm dt=\frac\pi{2\sqrt A}\big(1-\operatorname{erf}(a\sqrt A)\big)$$

$\endgroup$
0
0
$\begingroup$

Hint:

The factor $u$ can be seen as a term from the derivative of the exponent,

$$\frac{-2Au+B}{-2A},$$ which you easily integrate, giving a term $$\frac B{2A}.$$

Then the correction is the exponential of a quadratic polynomial, that you can turn to a Gaussian by a linear change of variable. As the lower bound is no more $0$, you need to resort to the antiderivative of a Gaussian, known as the function $\Phi$.

$\endgroup$

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.