I have to prove this identity: $$\int_{0}^{1}dx\int_{0}^{x}e^{x^2}dy=\int_{0}^{1}dy\int_{y}^{1}e^{x^2}dx$$ I've shown that: $$\int_{0}^{1}dx\int_{0}^{x}e^{x^2}dy=\int_{0}^{1}xe^{x^2}dx=\frac{1}{2}(e-1)$$
After i tried to solve the second member of identity in this way:
The first part is done by substituting $x^2$ in series $$ e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$ yielding $$e^{x^2} = \sum_{n=0}^{\infty} \frac{x^{2n}}{n!} $$ Through some theorems as uniform continuity, then we can switch the order of integration and summation, that is: $$\int_{y}^{1} \sum_{n=0}^{\infty} \frac{x^{2n}}{n!}dx= \sum_{n=0}^{\infty}\frac{1}{n!}\int_{y}^{1} x^{2n}dx=\sum_{n=0}^{\infty}\frac{1}{n!}*\Big(1-\frac{y^{2n+1}}{2n+1}\Big)dx$$ And from here i'm not sure how to proceed. What can i do? Thanks for the help in advance!!
Trying to calculate the iterated integrals is not the way here - one of those integrals will work in an elementary way, but the other won't.
No, this is a case of Fubini's theorem - the two integrals are the same because they're the same double integral over a triangle, integrated in the two possible orders. We wish to show that $$\int_0^1 \int_0^x f(x,y)\,dy\,dx = \int_0^1\int_y^1 f(x,y)\,dx\,dy$$ for nice enough $f$. The function we're trying to integrate is continuous and bounded on this bounded set, and that's certainly nice enough.
Draw the picture:
Integrating over $y$ first, our condition is that $0\le y\le x$. Integrating over $x$ first, our condition is that $y\le x\le 1$. Then, in both cases, the outer variable runs from $0$ to $1$.