Efficient Integration of Exponential Functions with Infinite Limits

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Homework Help Overview

The discussion revolves around the integration of the function (-ik - 2ax) * exp(-2ax^2) over the limits from negative to positive infinity. The subject area involves techniques of integration, particularly with exponential functions and infinite limits.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods for integrating the function, including the possibility of multiplying out the terms and using integration by substitution. There is also mention of evaluating the integral in parts and questioning the correctness of the approach taken.

Discussion Status

The discussion is ongoing, with participants exploring different methods for integration. Some guidance has been offered regarding separating the integral into parts and using substitution, but there is no explicit consensus on the best approach. Questions about the validity of certain methods and results are being raised.

Contextual Notes

Participants are navigating the complexities of integrating functions with infinite limits and are considering the implications of different integration techniques. There is a noted uncertainty about the correctness of specific approaches and results presented.

ronaldoshaky
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Homework Statement



how would you integrate, int (-ik - 2ax)*exp(-2ax^2) dx with limits infinity - infinity

Homework Equations



i think i can use the result int exp(-x^2) dx = sqrt (pi). But I am stumped.

The Attempt at a Solution



i thought about multiplying it out

int -ik*exp(-2ax^2) - 2ax*exp(-2ax^2) dx with limits infinity and - infinity

but it looks more complex now. would you use integration by substitution

Thank you
 
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Welcome to PF!

Hi ronaldoshaky! Welcome to PF! :smile:

(have an integral: ∫ and an infinity: ∞ and a pi: π and try using the X2 tag just above the Reply box :wink:)

Just do the two parts separately …

-∞ ik e-2ax2 dx you know how to do;

and ∫ 2ax e-2ax2 dx you can do by substitution. :wink:
 
Thank you!
 
Hi tiny-tim

Can I evaluate this integral ∫-∞∞ (ik-2ax) exp(-2ax2) dx without multiplying it out,
using integration by parts because it is a product?

I get


∫-∞∞ (ik-2ax) exp(-2ax2) dx = f(x)g(x) - ∫ f '(x)g(x) = - ik (sqrt (pi/2a)) + ikx (sqrt (pi/2a))

Is that any way correct?
 
No. :redface:
 

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