Question

given integral e^(-x^2) from -infinity to + infinity =pi^2 calculate the exact value of integral e^-(x-a)^2/b from -infinity to + infinity

Answer 1

\[\int\limits_{-\infty}^{\infty} e^(-x^2)dx = \sqrt{\pi}\] calculate \[\int\limits_{-\infty}^{\infty}e^-(x-a)^2/b\]

Answer 2

we clearly the -a has no effect, right?, because you can make a change of variables
u = x - a
and not introduce any constants.

Ok, now, make one more change of variables to get rid of the b in the exponent and what do you get?

Answer 3

The main question is what does pi have to do with the Gaussian function?

Answer 4

let me prove this statment

Answer 5

i know the answer don't rush?

Answer 6

That is not the main question. The pi is given and it is not asked to proved. I think you are going to confuse here by proving that relationship.

Answer 7

i am going to prove the whole mathematical concept

Answer 8

yes i know but the thing is if no one learns how to derive things then they will always forget the answer

Answer 9

if they don't understand the relationship then is this all about just getting the right answer to score a grade they don;t earn

Answer 10

i was tending to use the jacobians febnus method to prove the gaussian function

Answer 11

which uses matrix determinant and double integral, this proof was innovated by laplace

Answer 12

also polar coordinates

Answer 13

you can also use erf with limits to prove the concept

Answer 14

i'll leave this page but if required the proof let me know

Answer 15

@Ldaniel to typeset e^(expr), use e^{expr}
For example,
\[e^{-(x-a)^2}\] is \ [ e ^ { - (x - a ) ^ 2 } \ ] (without the spaces)

By default only the next character after the ^ symbol gets treated as an exponent unless you do that.

Answer 16

so how do I calculate the exact value? I don't know what to do with "b"

Answer 17

\[ \int\limits_{-\infty}^{\infty}e{ (x-a )^2}/b\]

Answer 18

\[\int\limits_{-\infty}^{\infty} e ^{-(u)^2}/b\] u=(x-a)

Answer 19

thanks by the way @whpalmer4

Answer 20

You're welcome. It was all I could do at that hour with the amount of blood in my caffeine stream :-)

Answer 21

\[\int\limits_{-\infty}^{\infty} e^ {(-(x-a)^2)/b} dx\] this this the right integral

Answer 22

hint:
\[\Large \frac{(x-a)^2}{b}=\left(\frac{x-a}{\sqrt {b}}\right)^2\]

Answer 23

to do this problem, use sirm's idea and do a change of variables.

let
\[ u^2 = \frac{(x-a)^2}{b} \]
\[ u = \frac{x}{\sqrt{b}}- \frac{a}{\sqrt{b}} \]

we need du in terms of dx (or vice versa). take the derivative of both sides, treating a and b as constants, and u and x as variables:
\[ du = \frac{1}{\sqrt{b}} dx \]
and, in terms of dx,
\[ \sqrt{b} \text{ }du = dx\]

we also need the limits for u. when x = -inf, we find
\[ u = \frac{-\infty}{\sqrt{b}}- \frac{a}{\sqrt{b}} = -\infty \]
similarly the upper limit for u is \( +\infty \)

now do the change of variables
\[ \int e^{-u^2} \sqrt{b} du = \sqrt{b} \int_{-\infty}^{+\infty} e^{-u^2} du \]
of course we could write this as
\[ \sqrt{b} \int_{-\infty}^{+\infty} e^{-x^2} dx \]

Answer 24

\[\sqrt{\pi b}\]

Answer 25

right?

Answer 26

yes

Answer 27

thank you so much @phi

Answer 28

The main thing is learn how to do the change of variables. It is a useful skill.

Answer 29

yeah