When evaluating the following integral (1) :
$$\int_0^v \int_0^{1} x^2e^{xy} \,dx\,dy$$
Using :
$$\frac{d}{dx} \int_a^{b} f(x,t)\,dt = \int_a^{b} \frac{∂}{∂x}f(x,t)\,dt
$$
I get:
$$ \int_0^{1}x^2e^{xy} \,dx = \frac{d^2}{dy^2}\int_0^{1}e^{xy}\,dx$$
Subbing this result into (1) gives:
$$\int \frac{d^2}{dy^2}[\frac{e^y}{y}-\frac{1}{y}]\,dy$$
Evaluating this integral gives: $$ \frac{e^y}{y}-\frac{e^y}{y^2}+\frac{1}{y^2}$$
What are the limits however of the last integral ???
PS: I Know there are other methods of obtaining the result (Integration by parts for example, you can even do the integral in reverse order.) However I want to find the answer using the above method . Is there a way to find the limits of the last integral without finding the inverse of the last function and if not, how does one find the inverse of the function?
The main issue actually is the fact I know the answer is not infinity, however plugging 0 straight into the last function yields us infinity. How does one actually find the limit of the last function as y --> 0
You have
$$ \lim_{y \to 0} \frac{ye^y - e^y -1}{y^2}. $$ Use l'hopotals to get $$ \lim_{y \to 0} \frac{ye^y + e^y -e^y}{2y} = \lim_{y \to 0} \frac{e^y}{2}= \frac12. $$