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November 21

Infinite Integral Sum

Hello! I'm a first year university student, and while practicing integration I made up a question $\int x^{2}e^{x^{2}}dx$ and tried to solve it using integration by parts. I got nowhere after a short while and, after looking the integral up, it turns out there is no elementary integral for the problem. However, integration by parts gives me this infinite series:

{\frac {1}{3}}x^{3}e^{x^{2}}-{\frac {2}{3}}\left({\frac {1}{5}}\right)x^{5}e^{x^{2}}+{\frac {2}{3}}\left({\frac {2}{5}}\right)\left({\frac {1}{7}}\right)x^{7}e^{x^{2}}-{\frac {2}{3}}\left({\frac {2}{5}}\right)\left({\frac {2}{7}}\right)\left({\frac {1}{9}}\right)x^{9}e^{x^{2}}+{\frac {2}{3}}\left({\frac {2}{5}}\right)\left({\frac {2}{7}}\right)\left({\frac {2}{9}}\right)\left({\frac {1}{11}}\right)x^{11}e^{x^{2}}\pm \cdots

I was just wondering, is there a way to write this expression as an infinite sum? I can kind of see a pattern but I don't have enough experience with infinite sums to generate one (if it even exists!). Thanks for your help. 70.54.112.243 (talk) 04:15, 21 November 2015 (UTC)[reply]

You just wrote down an infinite sum, so it is not clear to me what you want. You may put the first term outside parentheses to get the expression

{\frac {1}{3}}x^{3}e^{x^{2}}\sum _{i=0}^{\infty }\prod _{k=1}^{i}{\frac {-2x^{2}}{2k+3}}

Try also [[1]]

Bo Jacoby (talk) 05:59, 21 November 2015 (UTC).[reply]

This can also written without the product notation as

e^{x^{2}}\sum _{i=0}^{\infty }{\frac {(-1)^{i}4^{i+1}(i+2)!}{(2i+4)!}}x^{2i+3}

. -- Meni Rosenfeld (talk) 22:49, 21 November 2015 (UTC)[reply]

I'm sure you are familiar with Capital-sigma notation, but if you haven't seen Capital Pi notation before, you may wish to check out that article. (And while reading on the subject, you should familiarize yourself with Empty product.) Finaly, the erfi in the indefinite integral on the Wolfram Alpha page Bo Jacoby linked above is the imaginary error function. -- ToE 14:18, 21 November 2015 (UTC)[reply]

In J the function looks like this:

f=.13 : '3%~(y^3)*(^y^2)*+/*/\1,(-2*y^2)%3+2*1+i.10'

and a test run looks like this:

f&>0 0.5 1 1.5 2 2.5

0 0.0485128 0.627815 5.0843 46.7366 1398.4

Bo Jacoby (talk) 18:46, 22 November 2015 (UTC).[reply]

The Maclaurin series of this is straightforward to derive: $\int x^{2}e^{x^{2}}dx=\int x^{2}\sum _{i=0}^{\infty }{\frac {(x^{2})^{i}}{i!}}dx=\int \sum _{i=0}^{\infty }{\frac {x^{2i+2}}{i!}}dx=\sum _{i=0}^{\infty }{\frac {x^{2i+3}}{i!(2i+3)}}+C$ . Using the doubling formulas for the gamma function, this can be easily shown to be equivalent to the series you derived. --Jasper Deng (talk) 23:33, 23 November 2015 (UTC)[reply]