$Evaluate $I(a) = \int_{0}^{\infty}\frac{e^{-ax^2}-e^{-x^2}}{x}dx $$

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$Evaluate $I(a) = \int_{0}^{\infty}\frac{e^{-ax^2}-e^{-x^2}}{x}dx $$

I suspect that this problem involves Feynman's trick for integration but am unsure.

Evaluate $I(a) = \int_{0}^{\infty}\frac{e^{-ax^2}-e^{-x^2}}{x}dx $

Calculus

Curious Math Geek

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Please see attached. Thanks!

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Solution by Walter:

\[I(a)=\int_0^{\infty}\frac{e^{-ax^2}-e^{-x^2}}{x}.\]
Then
\[I'(a)=\int_0^{\infty}-x e^{-ax^2}=\frac{1}{2a} \int_0^{\infty}-2ax e^{-ax^2}dx=\frac{1}{2a} e^{-ax^2}|_0^{\infty}=-\frac{1}{2a}.\]
Thus
\[I'(a)=-\frac{1}{2a}.\]

Integrating with respect to $a$ over the interval $(1, a)$ we get
\[I(a)=-\frac{1}{2}\ln a\big |_{1}^{a}+I(1)=-\frac{1}{2}\ln a+\frac{1}{2}\ln 1+I(1).\]
We can easily see that
\[I(1)=\int_0^{\infty}\frac{e^{-x^2}-e^{-x^2}}{x}=0.\]
Hence
\[I(a)=-\frac{1}{2}\ln a.\]

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Kav10

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Walter

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@Kav10: You solution was incomplete and did not answer the question. I added an answer to the end of your solution. Please make sure that you can submit a complete solution before accepting similar questions in the future.
- Kav10
  
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  Thank you!
- Kav10
  
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  @Walter, your idea is good but I think you made a mistake in the differentiation, and that makes everything after that wrong! I have uploaded the correct differentiation and the rest of the solution.
Walter

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You are right, thank you for pointing that out. I revised my solution. You may want to remove the first solution you submitted as it may be confusing to Curious Math Geek.

$Evaluate $I(a) = \int_{0}^{\infty}\frac{e^{-ax^2}-e^{-x^2}}{x}dx $$

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