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I tried $u$-substiution with $u = x^2$ but that leaves me with $x$ values and not a simpler function to integrate.Is there a better way to integrate?

Michael Hardy
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    It can be proved that the antiderivative $x , \mapsto , \exp(x^2)$ cannot be expressed in terms of "elementary" functions. – pitchounet Oct 14 '14 at 22:12
  • Is your integration indefinite? Does it have any bound in integration? – CLAUDE Oct 14 '14 at 22:17
  • @AMIR No, but more specifically I'm trying to integrate $$x^5(e^(-x^2)$$ and integral $$3x^3(2^(x^2)$$ so I thought maybe integrating $$e^(x^2)$$ uses the same concepts. –  Oct 14 '14 at 22:18
  • put your mathematical expression in $$ in your comment to be readable – CLAUDE Oct 14 '14 at 22:20
  • @Ben You should open a new question stating the full problem. Someone will probably tell you to let $u=-x^2$ so that $x^4 x dx = -u^2\frac{1}{2}du$, or some such. This substitution reduces it to an integration by parts problem. – Mark McClure Oct 14 '14 at 22:22
  • unfortunately they are quite different in terms of difficulty. For $x^5e^{-x^2}$ you should use integration by parts and the integral of $e^{-x^2}$ (which is much easier, but still not exactly trivial). I reccomend making a new topic with the correct integrals. – JMoravitz Oct 14 '14 at 22:22
  • @Ben this is a famous integral known as the Gaussian integral that has no elementary antiderivatives and in fact, if the limits of integration are from negative to positive infinity, equals $\sqrt{\pi}$. – Sherlock Holmes Oct 15 '14 at 00:25

2 Answers2

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The result of integration can't be expressed using elementary functions. Such integrals can be calculated only on a defined segment and sometimes only numerically. In other words, there's no such function expressed as a trigonometrical, degree, exponential, logarithmic function, their sum, difference, multiplication, division or composition which derivative equals $e^{x^2}$.

Andrei Rykhalski
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Represent $e^{x^2}$ with the Maclauren series, then separate the term $\frac{1}{k!}$. You should now have an equation that can be easily integrated using the power rule. Integrate it and multiply the $\frac{1}{k!}$ term back in. We could stop here, but if you notice it equates to something close to the error function, then you can say that it is equal to the error function times other missing terms.

$$e^{x^2}=\sum_{k=0}^\infty\frac{(x^2)^k}{k!}\\ \int e^{x^2}=\sum_{k=0}^\infty\frac{1}{k!}\int x^{2k}dx\\ \int e^{x^2}=\sum_{k=0}^\infty\frac{x^{2k+1}}{k!(2k+1)}\\ \sum_{k=0}^\infty\frac{x^{2k+1}}{k!(2k+1)}=\int_0^x e^{-t^2}dt\\ erf(x)=\frac{2}{\sqrt\pi}\int_0^x e^{t^2}dt\\ \int e^{x^2}=-\frac{i\sqrt\pi}{2}erf(ix)$$

Anthony
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