Evaluate the Definite integral from zero to + infinite of (ax^2) * (e^-bx)

Question

I need to Evaluate the  Definite integral from zero to + infinite of (ax^2) * (e^-bx) . I don’t know how to integrate the e^-bx ( it’s e exposant -bx) and I need help to solve it. Thank you in advance !

REVISED QUESTION.

The expression should be (ax^2) * (e^-bx^2). See comments below.

Answer 1

REVISION COMPLETE!

Integrate by parts.

Let u=ax and dv=xe^-bx²dx, then du=adx and v=-e^-bx²/(2b).

 ∫udv=uv-∫vdu=-axe^-bx²/(2b)+(a/(2b))∫e^-bx²dx. The integral is an arbitrary Gaussian integral related to the error function erf. It evaluates to √(π/b) (the proof of this is lengthy and usually involves switching to polar coordinates, so this proof has been omitted) when taken over the whole real domain (-∞,∞). Because of symmetry, if the limits are [0,∞), then the integral (effectively the area beneath the positive half of the bell-shaped curve) is ½√(π/b). You can find details of the proof on-line in text and videos.

So (a/(2b))∫e^-bx²dx=(a/(2b))(½√(π/b))=a√(πb)/(4b²).

We still have to evaluate -axe^-bx²/(2b) between the same limits, which creates a problem, because, as x→∞, e^-bx²→0. However, the exponential factor is much more effective than the linear factor, so this term approaches zero quite rapidly. (When x=1, it's -a/e^b, when x=2, it's -2a/e^4b, so for large x the term becomes insignificant.).

Therefore the integral is a√(πb)/(4b²).

ADDENDUM

THE ARBITRARY GAUSSIAN INTEGRAL

Let I=∫e^-bx²dx, I²=[∫e^-bx²dx]²=[∫e^-bx²dx][∫e^-by²dy]=∬e^-b(x²+y²)dxdy. The limits of integration are (-∞,∞). x and y are placeholders, hence the odd appearance of the derivation of the double integral.

Let x=rcosθ, y=rsinθ, then r²=x²+y², tanθ=y/x. 

I²=∬e^-br²rdrdθ, where the limits are r=(-∞,∞) and θ=[0,2π]. 

But the function f(r)=re^-br²is an odd function, because f(-r)=-f(r), so the positive and negative areas beneath the curve for f(r) would cancel, that is, the integral over the domain (-∞,∞) would be zero. If we take the integral over [0,∞) and double it we would get the absolute area, the sum of the magnitudes of the positive and negative halves of f(r). So the intervals for r and θ would be [0,∞) and [0,π].

To evaluate the double integral start with the inner integral and let u=e^-br², so du=-2bre^-br²dr, re^-br²dr=-du/(2b).

Integrate this and we get -u/(2b)=-e^-br²/(2b), which after applying the limits [0,∞) becomes 0-(-1/(2b))=1/(2b).

The outer integral now becomes ∫dθ/2b=θ/(2b) for limits [0,π]=π/(2b)

Double this and we get π/b. I²=∬e^-br²rdrdθ, therefore I=√(π/b) or √(πb)/b when rationalised.

Comments

Thank you a lot for taking your time to answer , may I can ask you another question ? In fact I forgot to put a ^2 at the end, so the integral I need to make is ax^2 * e^-bx^2 and not the one that I wrote . When I tried it, I always have an error function and I’m trying to doing it without having the error function if it’s possible :))

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That makes a difference. So the answer needs to be revised. I can do that in due course, if you're prepared to wait.

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For sure I can wait there is no problem thank you a lot for taking your time it’s much appreciated!

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Revision is complete.

I have added an explanatory section in case you’re wondering how to integrate the error function (the special case of the Gaussian integral when b=1).