Expected value of X*exp(X) for X normally distributed

[TaPaKaH]

Assume we have $X\sim\exp(\mu,\sigma^2)$.
How does one compute $\mathbb{E}\left(Xe^X\right)$ and/or what is the outcome value?

 

[CompuChip]

The definition would say that

$$\mathbb{E}[ X e^X ] = \int_{-\infty}^{\infty} x e^{x} e^{-(x - \mu)^2 / \sigma^2} \, dx$$

That's a tricky one, I don't think any of the common integration strategies really work.
WolframAlpha does give me an exact result for the standard distribution $(\mu, \sigma) = (0, 1)$.

Maybe this will also help.

Sorry for the incomplete answer, hoping that this will get you started.

 

[D H]

Rewrite $\exp(x)\exp(-(x-\mu)^2/(2\sigma^2))$ as $\exp(x-(x-\mu)^2/(2\sigma^2)) = \exp((2\sigma^2x-(x-\mu)^2)/(2\sigma^2))$. Now complete the square so that you get $a\exp(-(x-\mu')^2)/(2\sigma^2))$, where $a$ is the exponential of the nasty junk you had to add to complete the square and $\mu'$ has the effect of being a new mean.

 

[CompuChip]

Very nice one, and congratulations on post number 13,000!