- #1

relatively_me

- 18

- 0

Int[e^(x^2) dx] limits->{-inf, +inf}

Thanks in advance

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- Thread starter relatively_me
- Start date

- #1

relatively_me

- 18

- 0

Int[e^(x^2) dx] limits->{-inf, +inf}

Thanks in advance

- #2

fahd

- 40

- 0

take x^2=t and proceed by parts...its fairly simple

- #3

relatively_me

- 18

- 0

thanks, i'll try that now

- #4

relatively_me

- 18

- 0

2x*e^(x^2) + x^2*e^(x^2)?

- #5

Pietjuh

- 76

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- #6

amcavoy

- 665

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Are you sure the problem wasn't with a negative xPietjuh said:

[tex]\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}[/tex]

Alex

- #7

himanshu121

- 654

- 1

You might try this link, it should be negative i believe, isn't there any typo

- #8

HallsofIvy

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fahd said:take x^2=t and proceed by parts...its fairly simple

I have no idea what you mean by that! What "parts" would you use? What would be u and what dv?

As others have pointed out, the infinite integral, as stated, does not exist. If that were [tex]e^{-x^2}[/tex] then it would be possible but certainly not "fairly simple"- there is no elementary anti-derivative.

- #9

Taviii

- 9

- 0

e^x^2=e^x*e^x

so

e^x=t

e^xdx=dt......replace in your exercesise and it should be easy from now on!

so

e^x=t

e^xdx=dt......replace in your exercesise and it should be easy from now on!

- #10

HallsofIvy

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[tex]\int (e^x)^2dx[/tex]

I guess I could do that- although I think I would be more likely to simply write

[tex](e^x)^2= e^{2x}[/tex]

and integrate that. HOWEVER, the integral asked about was

[tex]\int e^{x^2}dx[/tex]

and it is well known that that has no elementary anti-derivative.

- #11

1800bigk

- 42

- 0

if you are in analysis it can be solved using riemann-stieltjes or lebesgue theory, I saw this problem before and there was a theorem we used, it has slipped my memory.

it was something like if f(x) is differentiable on [a,b] and f prime is continuous on [a,b] then you can convert the integrand to somethin easier to integrate (i forget the rest) sorry

it was something like if f(x) is differentiable on [a,b] and f prime is continuous on [a,b] then you can convert the integrand to somethin easier to integrate (i forget the rest) sorry

Last edited:

- #12

what

- 91

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- #13

darkar

- 187

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what said:

I got this....sqrt(pi)Erfi(x)/2

What is that Erfi?

- #14

benorin

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Erfi is the imaginary error function above quoted by what (cool psuedonym).

- #15

BerkMath

- 60

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Taviii said:e^x^2=e^x*e^x

so

e^x=t

e^xdx=dt......replace in your exercesise and it should be easy from now on!

My guess is that Tavi will think things through before typing incorrect answers, next time.

- #16

HallsofIvy

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benorin said:Erfi is the imaginary error function above quoted by what (cool psuedonym).

All my errors are imaginary!

- #17

darkar

- 187

- 0

Umm..........

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