Hard integral (for me)

1. Jan 4, 2006

twoflower

Hi,

could you please help me in evaluating this integral?

$$\int 3e^{2x}\sqrt{x+1}\ dx$$

I got to this point during solving one ODE and have no idea what to do with this one.

Thank you.

2. Jan 4, 2006

benorin

not pretty, Maple says it involves an erfc.

3. Jan 4, 2006

twoflower

Bad, very bad...

4. Jan 4, 2006

fleon

Hi twoflower

I solve your integral, and found the following answer

$$\frac{3(4e^{2+2x}\sqrt{x+1}-\sqrt{2\pi}Erfi(\sqrt{2(1+x)}))}{8e^{2}}$$

when

Erfi[z] gives the imaginary error function erf(iz)/i

5. Jan 4, 2006

twoflower

Thank you, but that's definitely not what I was supposed to get..Kind of ugly for ODE solution, isn't it? :)

6. Jan 4, 2006

fargoth

try $$y=\sqrt{x+1}$$
and then pick $$4ye^{2y^2}$$ as dv and $$y$$ for u... (integration by parts

Last edited: Jan 4, 2006
7. Jan 4, 2006

benorin

Last edited: Jan 4, 2006
8. Jan 4, 2006

fleon

Sorry, I'm learning to use this editor, the answer is as follow

$$\frac{3(4e^{2+2x}\sqrt{x+1}-\sqrt{2\pi}Erfi(\sqrt{2(1+x)}))}{8e^{2}}$$.

9. Jan 4, 2006

fargoth

oh yeah, and i forgot, after the integration in parts i suggested above.. you know the trick for $$e^{x^2}$$ right?

10. Jan 4, 2006

twoflower

Thank you fargoth, gonna try that.

The ODE is

$$y'' + 2y' + y = 3e^{x}\sqrt{x+1}$$

11. Jan 4, 2006

benorin

is it $e^{2x}\mbox{ or }e^{x}$?

12. Jan 4, 2006

twoflower

It's $e^{x}$

13. Jan 4, 2006

benorin

Well, if it is either of the above, then your answer is just nasty (erfc, and all that). But if, perhaps, you had, say $$y^{\prime\prime}+2y^{\prime} + y= 3e^{-x}\sqrt{x+1}$$, then your answer is rather sweet, namely $$y(x)=c_{1}e^{-x}+c_{2}xe^{-x}+\frac{4}{5}e^{-x}(x+1)^{\frac{5}{2}}$$. In fact, if the exponent of e in the nonhomogeneous part is anything other than -1, erfc (computer aided educated guess).

Last edited: Jan 4, 2006
14. Jan 4, 2006

twoflower

You're an oracle, benorin! Professor wrote in wrongly, there has to be $e^{-x}$ in the original ODE.

15. Jan 4, 2006

saltydog

Why should the mildly messy format of the first solution discourage you? The ODE as first written is perfectly solvable even for the second solution (assuming you're using variation of parameter and B(x) is defined as above) with the second one:

$$A(x)=-3\int xe^{2x}\sqrt{x+1}$$

That's just me though.

Edit: Oh yea, I haven't (yet) checked this.

Edit2: Oh yea: equal rights for special functions. You know I have to send a dollar to Lurflurf each time I use that saying don't you?

Last edited: Jan 4, 2006
16. Jan 4, 2006

twoflower

Now I computed it and got another solution..

And the official one our proffesor has on his webpage is

$$y = \frac{3}{2} \log \left(x^2 + \sqrt{x^4 + 1}\right)e^{-x} -2(x+1)^{\frac{3}{2}}xe^{-x} + \alpha e^{-x} + \beta xe^{-x}$$

17. Jan 4, 2006

twoflower

What??

18. Jan 4, 2006

saltydog

Hey Twoflower, what's going on with this? Your initial ODE was:

$$y^{''}+2y^{'}+y=3e^x\sqrt{1+x}$$

Using variation of parameter, I get:

$$y(x)=A(x)e^{-x}+B(x)xe^{-x}+c_1e^{-x}+c_2xe^{-x}$$

with:

$$B(x)=\frac{3\left(4e^{2+2x}\sqrt{1+x}-\sqrt{2\pi}\text{Erfi}(\sqrt{2(1+x)})\right)}{8e^2}$$

$$A(x)=-\frac{3\left(4e^{2+2x}\sqrt{1+x}(4x-3)+7\sqrt{2\pi}\text{Erfi}(\sqrt{2(1+x)})\right)}{32e^2}$$

Back-substituting y(x) (via Mathematica) into the LHS of the ODE, I get the RHS.

When I back-substitute the "official solution" you posted above into the LHS of the ODE, I get an expression not even close to the RHS, minus x or not.

Last edited: Jan 4, 2006
19. Jan 5, 2006

saltydog

Really, just what is anyway:

$$\frac{d}{dx}\text{Erfi}[x]$$

and:

$$\frac{d^2}{dx^2}\text{Erfi}[x]$$

And also, I think we should ask Twoflower to solve this for us:

$$y^{''}+2y^{'}+y=3e^x\sqrt{1+x},\quad y(0)=0,\;y'(0)=1$$

not numerically neither: figure out what c1 and c2 are.

Last edited: Jan 5, 2006
20. Jan 7, 2006

twoflower

Well, this will definitely not appear in the test and I would probably not manage it.

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