Evaluating ∫x(e^x^3)dx: Tips and Tricks for Integrating e^x^3

[Mathmanman]

So I was trying to evaluate ∫x(e^x^3) dx
I know that you use integration by parts but something stumps me.

I don't know how to integrate e^x^3! I know how to integrate e^3x but not e^x^3!
Can someone please tell me how to even start on e^x^3?

 

[micromass]

You cannot find a solution to this integral in closed form using the usual elementary functions. You can prove this rigorously using the Liouville theorem.

You can find an expression of the integral using the incomplete Gamma function or the exponential integral.

 

[Mathmanman]

But my textbook solutions has an answer for this...
So originally, I was trying to solve the differential equation:
y' + 3x^3y = x
And then I found that the integrating factor was: e^∫3x^2 dx
= e^x^3

And my textbook said to use the integrating factor method.

 

[micromass]

http://www.wolframalpha.com/input/?i=int+x*e^(x^3)dx

 

[micromass]

About the differential equation you need to solve. It has no elementary solution. You need the imaginary error function.

http://www.wolframalpha.com/input/?i=y'+++3*x^3+*+y+=+x

 

[HallsofIvy]

But my textbook solutions has an answer for this...
So originally, I was trying to solve the differential equation:
y' + 3x^3y = x

I presume you mean it was y'+ 3x^2y= x

And then I found that the integrating factor was: e^∫3x^2 dx
= e^x^3

And my textbook said to use the integrating factor method.

Okay, multiplying both sides of the equation by $e^{x^3}$ gives
$(e^{x^3}y)'= xe^{x^3}$ and, integrating,
$e^{x^3}y= \int_a^x te^{t^3}dt$ where "a" gives the constant of integration.

From that $y(x)= e^{-x^3}\int_a^x te^{t^3}dt$

That integral cannot be done in terms of elementary functions so either leave the solution as that or use the gamma function as wolfram alpha suggests.

 

[SammyS]

But my textbook solutions has an answer for this...
So originally, I was trying to solve the differential equation:
y' + 3x^3y = x
And then I found that the integrating factor was: e^∫3x^2 dx
= e^x^3

Shouldn't that integrating factor be $\displaystyle \ e^{\left(\Large \int 3x^3\,dx\right)}\ ?$

 

[HallsofIvy]

Shouldn't that integrating factor be $\displaystyle \ e^{\left(\Large \int 3x^3\,dx\right)}\ ?$

Unless, as I suggested, the original differential equation was y'+ 3x^2y= x.

 

[STEMucator]

You could also use the Taylor expansion, if the integral is indeed $\int xe^{x^3} dx = \sum_{n=0}^{∞} \frac{1}{n!} \int x^{3n + 1} dx$.

EDIT: You could also find values by using $\int_{a}^{b}$ instead of $\int$.

 

[Mathmanman]

@Zondrina can you explain the taylor expansion method

 

[STEMucator]

Using a Taylor expansion for $e^x$, derive an expansion for $e^{x^3}$. Now find one for $xe^{x^3}$ and then integrate it as desired.

If you're interested in the reason, uniform continuity might be worth a read.