Question

Integral from infinity to negative infinity of Asin^2(x)+Bx^(-2) dx (all equal to one)

Answer 1

\[1 = \int\limits_{-\infty}^{\infty}Asin^2(\alpha x)+Bx^{2}dx\]

raw text:
1 = int_{-infty}^{infty}Asin^2(alpha x)+Bx^{2}dx

Answer 2

is it odd or even?

Answer 3

Sin(x) is odd: f(–x) = –f(x)
A squared function is even.

Answer 4

On the question box, you said Bx^(-2)
On comment box, you said Bx^2
Which one is the correct one?

Answer 5

Are you getting at something about the four conditions that need to be satisfied for a wave function? (1) Cont. (2) No corners or cusps (3) Finite area (4) squared integral (normalizable)

Answer 6

yes!!

Answer 7

And we have to prove it?

Answer 8

I need to find A & B.

Answer 9

\[1 = \int\limits_{-\infty}^{\infty}Asin^2(\alpha x)+Bx^{-2}dx\]

raw text:
1 = int_{-infty}^{infty}Asin^2(alpha x)+Bx^{-2}dx

Answer 10

if this is a wave function, you also have some IV's


are you presenting all of the question?!

Answer 11

A wavefunction for a particle confined in some region of space is given by the expression:
φ(x)=Asin^2(αx)+Bx^(-2)
What are the dimensions (i.e., units) of A and B?

Answer 12

lol!!

Answer 13

i think the wave eq is dimensionless

and so is A

B is oppo of 1/m^2

that's not yer question, though?!?!

Answer 14

Nope, that's not my question. I thought that's what I was suppose to do. Why are they opposite of that? And what is m?

Answer 15

ask something similar in the physics forum :-)

Answer 16

That's why I tried to turn it into a math problem (very few people available for the other subjects)

Answer 17

\[1 = \int\limits_{-\infty}^{\infty}Asin^2(\alpha x)+Bx^{-2}dx\]
Integrating something basically multiplies it by the dimension of integration.
So integrating the modulus multiplies the result by meters:
\[1 = \int\limits_{-\infty}^{\infty}1/(Asin^2(\alpha x)+Bx^{-2})dx\]

Right? the modulus??