Integral of a function with exponent

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  • #1
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Homework Statement



As part of a problem I need to evaluate,

[itex]\int[/itex][itex]^{\infty}_{0}[/itex]x2e-[itex]\lambda[/itex]x2

Homework Equations



∫e-λx2 = 1/2 [itex]\sqrt{\pi/\lambda}[/itex]
( with integration limits from 0 to ∞ )

The Attempt at a Solution


I've tried integrating by parts, also tried changing variables and combinations of these. been stuck on this integral for a couple of days, would appreciate some help.
 

Answers and Replies

  • #2
phyzguy
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What happened when you integrated by parts?
 
  • #3
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Homework Equations



∫e-λx2 = 1/2 [itex]\sqrt{\pi/\lambda}[/itex]
( with integration limits from 0 to ∞ )

The Attempt at a Solution


I've tried integrating by parts, also tried changing variables and combinations of these. been stuck on this integral for a couple of days, would appreciate some help.
when integrated by parts we have to integrate ∫e-λx2 dx as a function of x , but since its strictly increasing function of x i think it will diverge for the limit ∞
 
  • #5
phyzguy
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when integrated by parts we have to integrate ∫e-λx2 dx as a function of x , but since its strictly increasing function of x i think it will diverge for the limit ∞

That's not right. The function exp(-lambda x^2) decreases rapidly as x increases, so its integral from 0 to infinity is a known number. In fact you are given this value as part of the problem, no?
 
  • #6
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its decreasing if lambda is positive.

Your integral is the third integral down from this link. As to solving for it, you might have to do some research, but this will be a nice start for you.
thanks ! but i am talking about the function f(x) = [itex]\int e^{-λx^{2}}dx[/itex] when integrated from 0 to ∞

That's not right. The function exp(-lambda x^2) decreases rapidly as x increases, so its integral from 0 to infinity is a known number. In fact you are given this value as part of the problem, no?
what's the value of the indefinite integral [itex]\int e^{-λx^{2}}dx[/itex] ?
 
  • #7
HallsofIvy
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I don't see why integrating by parts wouldn't work. You did take [itex]u= x[/itex], [itex]dv= xe^{\lambda x^2}[/itex], didn't you?
 
  • #8
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I don't see why integrating by parts wouldn't work. You did take [itex]u= x[/itex], [itex]dv= xe^{\lambda x^2}[/itex], didn't you?
yea i got
 
  • #9
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Ok when I do by-parts, I get


-∫(2x).(∫e-λx2dx)dx + (x2)(∫e-λx2dx)|[itex]^{\infty}_{0}[/itex]

But I'm unsure how to proceed. I only know the integral ∫e-λx2dx when the limits are 0 and ∞, and I don't think I can simply apply the formula I posted. Even if I could, the function would be a constant and then i have both terms going to infinity.
 
  • #11
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Ok when I do by-parts, I get


-∫(2x).(∫e-λx2dx)dx + (x2)(∫e-λx2dx)|[itex]^{\infty}_{0}[/itex]

But I'm unsure how to proceed. I only know the integral ∫e-λx2dx when the limits are 0 and ∞, and I don't think I can simply apply the formula I posted. Even if I could, the function would be a constant and then i have both terms going to infinity.

I don't see why integrating by parts wouldn't work. You did take [itex]u= x[/itex], [itex]dv= xe^{\lambda x^2}[/itex], didn't you?

use what HallsofIvy posted. You will get the right answer.
 
  • #12
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ok i'll use u=x and dv = xe-λx2

Then I need to find v = ∫xe-λx2

for this I again integrate by parts

then v = -∫e-λx2 + x∫e-λx2 |[itex]^{\infty}_{0}[/itex]

The first term I know is 1/2([itex]\pi[/itex]/λ)1/2 but now the second term goes to infinity?
 
  • #13
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ok i'll use u=x and dv = xe-λx2

Then I need to find v = ∫xe-λx2

for this I again integrate by parts

nuh uhh. if u = x^2, du / dx = 2x. dx = du/2x, getting rid of one of the x's
 
  • #14
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The first term I know is 1/2([itex]\pi[/itex]/λ)1/2 but now the second term goes to infinity?

that second integral definitely converges...as I think was stated earlier in this thread a couple of times.

There is nothing wrong with using IBP here, but it's not necessary if you look at my above post.
 
  • #15
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Ah I see, so now I get

v = -(1/2λ)e-λx2

and my integral becomes

(1/2λ)∫[itex]^{\infty}_{0}[/itex]e-λx2dx - x(1/2λ)e-λx2 |[itex]^{∞}_{0}[/itex]

and the second term goes to zero as x tends to ∞ because of the decaying exponent term.

giving my intergral as (1/4λ) ([itex]\frac{\pi}{\lambda}[/itex])1/2

Is that right?
 
  • #16
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Is that right?

If I remember correctly from that wikipedia link I sent you, yes.
 
  • #17
SammyS
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ok i'll use u=x and dv = xe-λx2

Then I need to find v = ∫xe-λx2dx

for this I again integrate by parts

then v = -∫e-λx2 + x∫e-λx2 |[itex]^{\infty}_{0}[/itex]

The first term I know is 1/2([itex]\pi[/itex]/λ)1/2 but now the second term goes to infinity?
For the integral in the equation for v, [itex]v=\int xe^{-\lambda x^2}dx\,,[/itex] you don't need to use integration by parts. Instead, use substitution. Let u = -λx2 → du = -2λ x dx .
 
  • #18
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thanks a lot, you've been a big help
 

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