The power series for e^xex is sum_(n=0)^oox^n/(n!)∞∑n=0xnn!. Substituting in x^2x2 for xx, we get
e^(x^2) = sum_(n=0)^oo(x^2)^n/(n!) = sum_(n=0)^oox^(2n)/(n!)ex2=∞∑n=0(x2)nn!=∞∑n=0x2nn!
Substituting this in, we have
int_0^1e^(x^2)dx = int_0^1sum_(n=0)^oox^(2n)/(n!)dx = sum_(n=0)^ooint_0^1x^(2n)/(n!)dx∫10ex2dx=∫10∞∑n=0x2nn!dx=∞∑n=0∫10x2nn!dx
where the second equality is valid due to x^(2n)/(n!) >= 0x2nn!≥0 for all x, nx,n.
Evaluating the new integral, we have
sum_(n=0)^ooint_0^1x^(2n)/(n!)dx = sum_(n=0)^oo1/(n!)int_0^1x^(2n)dx∞∑n=0∫10x2nn!dx=∞∑n=01n!∫10x2ndx
=sum_(n=0)^oo1/(n!)[x^(2n+1)/(2n+1)]_0^1=∞∑n=01n![x2n+12n+1]10
=sum_(n=0)^oo1/(n!(2n+1))=∞∑n=01n!(2n+1)
At this point we only need to calculate enough terms of the above sum to get an answer accurate to 44 decimal places. We could either keep calculating them until it becomes obvious, or do a little more work to get a bound on how many terms we need. In either case, calculating enough terms shows us that
sum_(n=0)^oo1/(n!(2n+1)) = 1.46265... ~~1.4627
