Compute $\mathbb{E}\big[\exp(XY+X)\big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.

Question

Compute $\mathbb{E}\big[\exp(XY+X)\big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.

I first computed $\mathbb{E}\big[\exp(XY+X)\mid X\big]$.

Because $X$ and $Y$ are independent then $$\mathbb{E}\big[\exp(XY+X)\mid X\big] = \phi(X)\,,$$ where $$\phi(x) =\mathbb{E}\big[\exp(xY+x)\big] = \int_0^1 \exp(xy+x)dy = \frac{\exp x(\exp x - 1)}{x}.$$ so $$\mathbb{E}\big[\exp(XY+X)\big] = \int_0^1 \frac{\exp x(\exp x - 1)}{x} dx\,,$$ which I don't know how to compute.

Could someone check if I'm good till now and maybe help me continue? Thanks!

Answer 1

You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $\operatorname{Ei}$ is given by $$\operatorname{Ei}(t)=\mathrel{-\!\!\!\!\!\!\;\!\!\int}_{-\infty}^{t}\frac{e^{s}}{s}ds.$$ (Here $\mathrel{-\!\!\!\!\!\;\!\!\int}$ is the Cauchy principal value integral.) In other words, $$\int\frac{e^{s}}{s}ds=\operatorname{Ei}(s)+C.$$ This also shows that $$\int\frac{e^{ks}}{s}ds=\operatorname{Ei}(ks)+C.$$

The required integral is \begin{align}I&=\int_0^1\frac{e^x(e^x-1)}{x}dx=\lim_{\epsilon\searrow0}\int_{\epsilon}^1\left(\frac{e^{2x}}{x}dx-\int_\epsilon^1\frac{e^x}{x}dx\right)\\ &=\lim_{\epsilon\searrow0}\Big(\big(\operatorname{Ei}(2)-\operatorname{Ei}(2\epsilon)\big)-\big(\operatorname{Ei}(1)-\operatorname{Ei}(\epsilon)\big)\Big)\\ &=\operatorname{Ei}(2)-\operatorname{Ei}(1)-\lim_{\epsilon\searrow0}\big(\operatorname{Ei}(2\epsilon)-\operatorname{Ei}(\epsilon)\big).\end{align} Now $$\operatorname{Ei}(2\epsilon)-\operatorname{Ei}(\epsilon)=\int_{\epsilon}^{2\epsilon}\frac{e^s}{s}ds=\int_\epsilon^{2\epsilon}\frac{1+O(\epsilon)}{s}ds=\int_{\epsilon}^{2\epsilon}\frac{ds}{s}+O(\epsilon)=\ln 2+O(\epsilon).$$ This gives $$I=\operatorname{Ei}(2)-\operatorname{Ei}(1)-\ln2\approx 2.366.$$

Answer 2

You can't get an explicit answer. However let $u=\exp(x)$ then $du=\exp(x)dx$ and your integral becomes $\int_1^e \frac{u-1}{\ln(u)}du=\text{li}(e^2)-\text{li}(e)=\int_e^{e^2}\frac{1}{\ln(u)}du$. Unfortunately there is no explicit formula for $\text{li}(u)$. I calculated this result and got I=3.0591165.
In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $\lim_{\epsilon\to 0}\int_{(1+\epsilon)}^{(1+\epsilon)^2}\frac{1}{ln(u)}du=ln(2)=0.693147181$
Net result$=2.365969319$.