2
$\begingroup$

I am trying to calculate the following integral

$$ \int_{-\infty}^{\infty}\frac{\lambda_{1}}{2}e^{-\lambda_{1}|x|-\frac{\lambda_{2}}{2}x^{2}}dx $$

After some simplification I got

$$ 2\int_{0}^{\infty}\frac{\lambda_{1}}{2}e^{-\lambda_{1} x-\frac{\lambda_{2}}{2}x^{2}}dx $$

Does anyone know how to continue?

Also, can this be solved using the gamma function?

Thanks.

$\endgroup$
2
  • $\begingroup$ Look into the so-called "error function" instead (erf(x)=$\int_0^x \exp(-t^2)dt$). I think it has a better name, but this is a common one. $\endgroup$
    – Vincent
    Aug 26 '16 at 14:08
  • $\begingroup$ Since OP needs the definite integral, it's easier than that. $\endgroup$ Aug 26 '16 at 14:08
3
$\begingroup$

Note that we can write

$$\begin{align} I(\lambda_1,\lambda_2)&=\frac{\lambda_1}{2}\int_{-\infty}^\infty e^{-\lambda_1 |x|-\frac12 \lambda_2 x^2}\,dx\\\\ &=\frac{\lambda_1}{2}\int_{-\infty}^\infty e^{-\frac12 \lambda_2 \left(x^2+2\frac{\lambda_1}{\lambda_2}|x|\right)}\,dx\\\\ &=\frac{\lambda_1}{2}e^{ \lambda_1^2/2\lambda_2}\int_{-\infty}^\infty e^{-\frac12 \lambda_2 \left(|x|+\frac{\lambda_1}{\lambda_2}\right)^2}\,dx\\\\ &=\lambda_1 e^{ \lambda_1^2/2\lambda_2}\int_0^\infty e^{-\frac12 \lambda_2 \left(x+\frac{\lambda_1}{\lambda_2}\right)^2}\,dx\\\\ &=\lambda_1 e^{ \lambda_1^2/2\lambda_2}\int_{\lambda_1/\lambda_2}^\infty e^{-\frac12 \lambda_2 x^2}\,dx\\\\ &=\lambda_1 \sqrt{\frac{2}{\lambda_2}}e^{ \lambda_1^2/2\lambda_2}\int_{\lambda_1/\sqrt{2\lambda_2}}^\infty e^{-x^2}\,dx\\\\ &= \sqrt{\frac{\pi \lambda_1^2}{2\lambda_2}}e^{ \lambda_1^2/2\lambda_2}\text{erfc}\left(\sqrt{\frac{\lambda_1^2}{2\lambda_2}}\right) \end{align}$$

where $\text{erfc}(x)$ is the complementary error function.

$\endgroup$
6
  • $\begingroup$ You have found the right way (+1) ! I am in the process of writing another answer using Fourier Transform. $\endgroup$
    – Jean Marie
    Aug 26 '16 at 14:34
  • $\begingroup$ In the second-to-last equation you changed the integration limits but not the function to integrate. By the way, I had employed the simmetry of the integrand before completing the square: it is not straightforward that you can complete the square with an absolute value inside. $\endgroup$
    – N74
    Aug 26 '16 at 14:38
  • $\begingroup$ @N74 No, Dr. MV is right, if you shift variable by $\lambda_1/\lambda_2$, you have an integral with $|x|^2$ which is equal to $x^2$. $\endgroup$
    – Jean Marie
    Aug 26 '16 at 14:47
  • $\begingroup$ A different point (from 2nd-to-last equation into last one) I find $ \int_{\lambda_1/\sqrt{2\lambda_2}}^\infty e^{-\frac12 \lambda_2 x^2}\,dx= K \text{erfc}\frac{\lambda_1}{2\lambda_2}$ (for a certain constant $K$) in particular without square root. $\endgroup$
    – Jean Marie
    Aug 26 '16 at 14:52
  • $\begingroup$ @JeanMarie the error is in the second-to-last equation. There's no error in the absolute value handling, but it is harder to follow. $\endgroup$
    – N74
    Aug 26 '16 at 14:52
0
$\begingroup$

Using Dr. MV solution, as the OP asked for it, this is a solution with the incomplete gamma function. $$I(\lambda_1,\lambda_2)= \sqrt{\frac{ \pi \lambda_1^2}{2\lambda_2}}e^{ \lambda_1^2/2\lambda_2} -\sqrt{\frac{ \lambda_1^2}{2\lambda_2}}e^{ \lambda_1^2/2\lambda_2}\gamma\left({1\over 2},\frac{\lambda_1^2}{2\lambda_2}\right)$$

$\endgroup$
0
$\begingroup$

Defining Fourier Transform (FT) by the following formula:

$$\tag{1}\hat f(u):=\int_{-\infty}^{\infty}e^{2i\pi t u}f(t)dt$$

one has the isometry formula:

$$\tag{2}\int_{-\infty}^{\infty}f(t) g(t) dt \ = \ \int_{-\infty}^{\infty}\hat f(u) \hat g(u) du$$

Let us calculate separately (or adapt from tables because they are "avatars" of classical transform pairs) the FT of

  • $f(t):=\frac{\lambda_{1}}{2}e^{-\lambda_{1} |t|}$ is $\hat f(u)=\dfrac{\lambda_1^2}{\lambda_1^2+4 \pi^2 u^2}$ ("a symmetric exponential (Laplace dist.) is exchanged with a Cauchy function").

  • $g(t):=e^{-\frac{\lambda_2}{2}t^2}$ is $\hat g(u)=\sqrt{\dfrac{2\pi}{\lambda_2}}e^{-\frac{2 \pi^2}{\lambda_2}u^{2}}$ ("a peaky Gaussian function is tranformed into a flat Gaussian function")

It suffices now to apply (2) to obtain:

$$\int_{-\infty}^{+\infty}\dfrac{\lambda_1^2}{\lambda_1^2+4 \pi^2 u^2}\sqrt{\dfrac{2\pi}{\lambda_2}}e^{-\frac{2 \pi^2}{\lambda_2}u^{2}}du=\sqrt{\dfrac{2\pi}{\lambda_2}}\int_{-\infty}^{+\infty}\dfrac{1}{1+(K u)^2}e^{-\frac{1}{2 \sigma^2}u^{2}}du$$

where $K:=\dfrac{2 \pi}{\lambda_1} \ \ \text{and} \ \ \sigma:=\dfrac{\lambda_2}{2 \pi}.$

Expanding into series $\dfrac{1}{1+(K u)^2}=\sum_{n=0}^{\infty}(-1)^n(Ku)^{2n}$, integrating term by term by using the classical moments of the normal distribution ($\hat{m}_{2n} = \sigma^{2n} (2n-1)!!$), one recognizes the development of the complementary error function:

$$\sqrt{\pi}\lambda_3 \text{erfc}(\lambda_3) \ \ \text{where} \ \ \lambda_3:=\dfrac{\lambda_1}{\sqrt{2 \lambda_2}}$$

Remark: I have done at first this computation with Mathematica...

$\endgroup$

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.