Integrating absolute values over infinity

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SUMMARY

The discussion focuses on calculating the expectation value in terms of the constant X0 using the wave function Ψ(x) = (1/√X0)e^(-|x|/X0). The integral = (1/X0)∫(−∞ to ∞) Ψ* x Ψ dx simplifies to = (1/X0)∫(−∞ to ∞) e^(-2|x|/X0) dx. Participants clarify that the derivative of |x| is piecewise, with x' = -1 for x < 0 and x' = 1 for x > 0, and suggest that the integral can be evaluated by recognizing the even nature of the function, allowing integration from 0 to ∞ and doubling the result.

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ElijahRockers
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Homework Statement



Find <x> in terms of X0 if X0 is constant and

\Psi(x) = \frac{1}{\sqrt{X_0}}e^{\frac{-|x|}{X_0}}

and

&lt;x&gt; = \int^{\infty}_{-\infty}{\Psi^* x \Psi}dx

where Psi* is the complex conjugate of Psi.

Since there is no imaginary component, this is effectively Psi2.

so, from here I could do a u-substitution to integrate over e^u du, but I'm not sure how.

What is the derivative of -2|x|/X_0 with respect to x?

This is part of a physics exercise I'm working on.

&lt;x&gt; = \frac{1}{x_0}\int^{\infty}_{-\infty}e^{\frac{-2|x|}{X_0}} dx

I have found that the derivative of |x| depends on whether x<0 or x>0. for x<0, x'=-1 and for x>0, x'=1 but I'm not sure how to tie this all together for the integration.

I guess what I'm really asking is how do I find the integrand here?
 
Last edited:
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ElijahRockers said:

Homework Statement



Find <x> in terms of X0 if X0 is constant and

\Psi(x) = \frac{1}{\sqrt{X_0}}e^{\frac{-|x|}{X_0}}

and

&lt;x&gt; = \int^{\infty}_{-\infty}{\Psi^* x \Psi}dx

where Psi* is the complex conjugate of Psi.

Since there is no imaginary component, this is effectively Psi2.

so, from here I could do a u-substitution to integrate over e^u du, but I'm not sure how.

What is the derivative of -2|x|/X_0 with respect to x?

This is part of a physics exercise I'm working on.

&lt;x&gt; = \frac{1}{x_0}\int^{\infty}_{-\infty}e^{\frac{-2|x|}{X_0}} dx

I have found that the derivative of |x| depends on whether x<0 or x>0. for x<0, x'=-1 and for x>0, x'=1 but I'm not sure how to tie this all together for the integration.

I guess what I'm really asking is how do I find the integrand here?
So, split the integral.

\displaystyle &lt;x&gt; = \frac{1}{x_0}\int^{0}_{-\infty}xe^{\frac{-2|x|}{X_0}} dx+\frac{1}{x_0}\int^{\infty}_{0}xe^{\frac{-2|x|}{X_0}} dxAdded in Edit:

And where did the x go in the integrand ? Well, that makes the integrand odd. Therefore, don't bother splitting it.
 
Last edited:
ElijahRockers said:

Homework Statement



Find <x> in terms of X0 if X0 is constant and

\Psi(x) = \frac{1}{\sqrt{X_0}}e^{\frac{-|x|}{X_0}}

and

&lt;x&gt; = \int^{\infty}_{-\infty}{\Psi^* x \Psi}dx

where Psi* is the complex conjugate of Psi.

Since there is no imaginary component, this is effectively Psi2.

so, from here I could do a u-substitution to integrate over e^u du, but I'm not sure how.

What is the derivative of -2|x|/X_0 with respect to x?

This is part of a physics exercise I'm working on.

&lt;x&gt; = \frac{1}{x_0}\int^{\infty}_{-\infty}e^{\frac{-2|x|}{X_0}} dx

I have found that the derivative of |x| depends on whether x<0 or x>0. for x<0, x'=-1 and for x>0, x'=1 but I'm not sure how to tie this all together for the integration.

I guess what I'm really asking is how do I find the integrand here?

Just integrate from 0 to infinity where |x|=x. The integral from -infinity to 0 will be the same thing since your function is even.
 

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