Integral of exp function?

In summary, you are meant to integrate e^{-a^2x^2}/a^2 but instead get e^{-a^2x^2}/a^2. This incorrect result can be easily verified by taking the derivative.
  • #1
philip041
107
0
I am meant to get the integral

[tex]
\int^{\infty}_{-\infty}{dxe^{-\alpha^2x^2}}
[/tex]

to equal

[tex] A^2\frac{\sqrt{\pi}}{\alpha} [/tex]

but I get

[tex]
A^2\left[\frac{e^{-\alpha^2x^2}}{\alpha^2}\right]^{\infty}_{-\infty}
[/tex]

which I'm pretty sure doesn't equal much.

Thoughts?
 
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  • #2
Argh I don't know how to sort the Latex out, suggest how and I will edit it, or if you want repost my stuff. It works in my Latex I don;t know what I'm doing wrong!
 
  • #3
The slash should be the other way around like this:

[tex]
\int^{\infty}_{-\infty}{e^{-\alpha^2x^2}} \mbox{d}x
[/tex]
 
  • #4
philip041 said:
I am meant to get the integral

[tex]
\int^{\infty}_{-\infty}{dxe^{-\alpha^2x^2}}
[/tex]

to equal

[tex] A^2\frac{\sqrt{\pi}}{\alpha} [/tex]

but I get

[tex]
A^2\left[\frac{e^{-\alpha^2x^2}}{\alpha^2}\right]^{\infty}_{-\infty}
[/tex]

which I'm pretty sure doesn't equal much.

Thoughts?
Your closing tex tags use a backslash but should use a forward slash. I have changed them in what I've quoted from you.

Regarding the integral, you don't show how you got what you show. Inasmuch as this is an improper integral, you probably should have split this into two integrals and used limits.

Also, you seem to have gotten [tex]e^{-a^2x^2}/a^2[/tex] as your antiderivative, which is incorrect. To see that this is so, take the derivative.
 
  • #5
That's a fairly standard calculation but because the anti-derivative is NOT an "elementary" function, requires a couple of tricks to do that definite integral.

You want to find
[tex]I= \int_{-\infty}^{\infty} e^{-\alpha^2x^2}dx[/tex]

First, because the integrand is symmetric about x= 0,
[tex]\frac{I}{2}= \int_0^\infty e^{-\alpha^2x^2}dx[/itex]

It is also true that
[tex]\frac{I}{2}= \int_0^\infty e^{-\alpha^2y^2}dy[/itex]
so that
[tex]\frac{I^2}{4}= \left(\int_0^\infty e^{-\alpha^2x^2}dx\right)\left(\int_0^\infty e^{-\alpha^2y^2}dy\right)[/tex]
[tex]= \int_{R^+\times R^+}\int e^{-\alpha^2(x^2+ y^2)}dA[/tex]
Where the integral is over the first quadrant. Write that in polar coordinates and it becomes easy to integrate.
 
  • #6
Cheers
 

What is the integral of e^x?

The integral of e^x is equal to e^x + C, where C is a constant. This can be written as ∫ e^x dx = e^x + C.

How do you solve an integral of e^x?

To solve an integral of e^x, you can use the formula ∫ e^x dx = e^x + C, where C is a constant. You can also use integration by parts or substitution method.

What is the relationship between the integral of e^x and its derivative?

The integral of e^x is the inverse operation of the derivative of e^x. In other words, the derivative of e^x is e^x, and the integral of e^x is e^x + C.

Is there a specific method for solving integrals of exponential functions?

Yes, there are specific methods for solving integrals of exponential functions, such as the substitution method, integration by parts, or using the formula ∫ e^x dx = e^x + C. It is important to choose the appropriate method based on the complexity of the integral.

Can the integral of e^x be solved using a calculator?

Yes, most scientific calculators have a built-in function for solving integrals. However, it is still important to understand the concepts and methods for solving integrals, rather than relying solely on a calculator.

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