Question

is integral from 1 to inf of e^x^2 convergent or divergent

Answer 1

Hmmm...

Answer 2

I remember you can do something like this...

Answer 3

\[\Large\rm \int\limits_1^{\infty} e^{x^2}~dx \cdot \int\limits_1^{\infty} e^{y^2}~dx =\int\limits \int\limits e^{x^2+y^2}~dx~dy\]Changing to polar,\[\Large\rm x=r\cos \theta\]\[\Large\rm y=r \sin \theta\]Changing our integral to,\[\Large\rm \int\limits_{\theta}\int\limits_{r} e^{r^2}(r dr~d \theta)\]Making it a lot easier to integrate.
Buuuuuut the boundaries, hmmm....

Answer 4

Oh oh and keep in mind that I multiplied two of the initial integrals together.
So if we're calling that integral \(\Large\rm I\), then the first step was \(\Large\rm I^2\).

So after doing all the integration, we would need to take the square root of the result.

Answer 5

hmmm we havn't learned polar coordinates yet... -___- lol is there another way?

Answer 6

you may use comparison test, look for some converging function f(x) thats greater than e^x^2 in the given integral

Answer 7

what could my comparison function be?

Answer 8

simply e^x?

Answer 9

Notice that \(\large e^{x^2}\) shoots up quickly and so diverges for sure,
are you really sure you're not missing a negative sign in the exponent ?

Answer 10

Oh my bad :( They didn't want the actual integral value, they just wanted convergent divergent :3 hehe woops.

Answer 11

loll its ok i appreciate all that time you spent making such nice equations :D

Answer 12

i am not missing anything, although our professor has been known to accidentally give us problems that are too difficult and he might change them when someone askes in class (he writes all his own problems -__-)

Answer 13

if its really e^x^2, then its easy.. just use e^x as your comparison function

Answer 14

i dont know if it diverges for sure, i would have that gabriels horn was divergent by looking at the graph but it equals pi -__-

Answer 15

really? ok and then use LCTT?

Answer 16

you may want to show the work for evaluating the integral e^x

Answer 17

isn't e^x^2 bigger than e^x?

Answer 18

yeah it was a typo sorry to confuse u ;p

Answer 19

loll ok ;p thank you!

Answer 20

yay so direct comparison works!

Answer 21

thank you both :D

Answer 22

corrected typo :

\(\large e^{x^2} \gt e^x\) in the given interval
since the smaller function \(\large \int\limits_{1}^{\infty}e^x dx\) diverges, the integral in question also diverges

Answer 23

k thank you !