Show that $f(x,y) = \frac{e^x+1}{x}$ is Lebesgue integrable. Show that $$ f: [0,1)^2 \rightarrow \mathbb{R}, \quad f(x,y)=\left\{\begin{array}{cl} \frac{e^x+1}{x}, & \mbox{for }y \le x\\ 0, & \mbox{else} \end{array}\right. $$ is Lebesgue integrable and calculate $\int f d\lambda ^2$.
It would be nice if someone could help me with the above mentioned task. Im not really firm in calculating Lebesgue-Integrals that arent equal to Riemann-Integrals. Also, i know no real criteria for lebesgue integrability, other than that the improper riemann integral of $|f|$ is finite. Hints are also very welcome. 
 A: Since $x$ and $y$ are in $[0,1)$ and your function is continuous on each variable  then you can use the improper  Rieamnn integral to compute $|f|$ or $f$ since it is positive.  More precisely, let us fix $y$ in $[0,1)$ and consider $f$ as a function in $x$ in the interval $[y,1)$ then $$\int_{y}^{1} f(x)dx=\int_{y}^{1}\frac{e^{x}+1}{x} dx$$
$$\leq \int_{y}^{1} \frac{e+1}{x}=(e+1)\log(1/y)<\infty.$$
Then the improper Riemann integral of $|f|$ is finite which implies that this function is Lebesgue integrable. Now, let us compute $\int f d\lambda^2.$ Since $d\lambda^2=dx dy.$ We have $$\int f d\lambda^2=\int_{[0,1)^2} f(x,y) dx dy$$
$$=\int_{0}^{1} \int_{y}^{1} f(x,y) dx dy=\int_{0}^{1} \left[\int_{y}^{1} \frac{e^{x}+1}{x}dx\right] dy.$$ Put $v=e^{x}$ we get $dx=\frac{dv}{v}$ then
$$\int_{y}^{1} \frac{e^{x}+1}{x}dx =\int_{e^{y}}^{e} \frac{v+1}{v \log{v}} dv.$$
Now, integration by parts, with $u(v)=v+1$ and $w(v)=\frac{1}{v\log{v}}$ gives 
$$\int_{e^{y}}^{e} \frac{v+1}{v \log{v}} dv=\left[(v+1) \log(\log{v})\right]_{e^{y}}^{e}-\int_{e^{y}}^{e} \log(\log{v}) dv=-(e^{y}+1)\times \log{y}-\int_{e^{y}}^{e} \log(\log{v}) dv.$$
After computing the integral $g(y)=\int_{e^{y}}^{e} \log(\log{v}) dv$ which is a function in $y$ you finally compute the integral $\int_{0}^{1}(e^{y}+1)\cdot \log{y}-g(y)dy.$ 
