Definite integral with exponential

In summary, the conversation discusses the use of polar coordinates to evaluate an integral and the computation of an integral using a change of coordinates similar to the first part. The conversation also includes the use of substitutions to simplify the integrand and solving for a suitable substitution.
  • #1
Benny
584
0
Hi I have a few questions regarding the following two part problem.

a) Let a > 0. Use polar coordinates to evaluate [tex]\int\limits_{ - \infty }^\infty {e^{ - ax^2 + bx} } dx[/tex]

[tex]Ans:e^{\frac{{b^2 }}{{4a}}} \sqrt {\frac{\pi }{a}} [/tex]

b) Let [tex]\mathop x\limits^ \to = \left( {x_1 ,...,x_n } \right) \in R^n [/tex], [tex]\mathop b\limits^ \to \in R^n [/tex] and [tex]A = U^T U[/tex] where U is an invertible n * n matrix. Compute

[tex]\int\limits_{R^n }^{} {\exp \left( { - x^T Ax + b^T x} \right)} dx_1 ...dx_n [/tex]

(the x and b in the argument of the exponential are vectors)

Check that your answer for n = 1 reduces to the answer in (a).

Ans: [tex]Ans:\frac{{\pi ^{\frac{n}{2}} }}{{\left| {\det U} \right|}}\exp \left( {\frac{1}{4}b^T A^{ - 1} b} \right)[/tex]


My questions/work done etc. are as follows.

In part (a) I think the idea is to make a substitution which leaves me with an integrand which is a scalar multiple of exp(-y^2). This is so that I can then use the standard 'squaring the integral' method to get the answer.

I had a few problems doing this. Basically after trying a few different methods I just subtituted [tex]x = \frac{1}{{\sqrt a }}\left( {y - c} \right)[/tex] (c is constant) into -ax^2 + bx. Then I selected c so that the coefficient of the resulting y term is zero. I'm wondering if there is a simpler way to go about finding a suitable substitution for the integral.

I don't follow the solution to part b. They start out by saying that a change of coordinates similar to part a needs to be used and come up with [tex]x = U^{ - 1} \left( {y + \frac{1}{2}U^{ - 1} b} \right)[/tex]. I don't understand where they got it from. Could someone suggest how they would start this question? Any help would be good thanks.
 
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  • #2
Benny said:
In part (a) I think the idea is to make a substitution which leaves me with an integrand which is a scalar multiple of exp(-y^2). This is so that I can then use the standard 'squaring the integral' method to get the answer.
I had a few problems doing this. Basically after trying a few different methods I just subtituted [tex]x = \frac{1}{{\sqrt a }}\left( {y - c} \right)[/tex] (c is constant) into -ax^2 + bx. Then I selected c so that the coefficient of the resulting y term is zero. I'm wondering if there is a simpler way to go about finding a suitable substitution for the integral.

Yes, first complete the square in the exponent to arrive at something like:

[tex]e^{-a(x+k)^2}[/tex]

Then "scale out" the a by using another substitution:

[tex] \sigma=\sqrt{a}x[/tex]

. . . thanks Tide.:smile:
 
  • #3
Thanks for the help Saltydog.

There's another thing related to my original post which I am unsure of.

In part b they used the substitution: [tex]x = U^{ - 1} \left( {y + \frac{1}{2}\left( {U^T } \right)^{ - 1} b} \right)[/tex] where x, U, b etc are as defined in my initial post.

[tex]
x^T = \left( {y^T + \frac{1}{2}b^T U^{ - 1} } \right)\left( {U^T } \right)^{ - 1}
[/tex]

I understand how most of the transpose of x (as given above) is obtained. However, I can't see how the bracket with the y term in x^T can contain U^-1.

Using properties of the transpose to find an expression for x^T, there would be a step where I need to take the transpose of (U^T)^-1. Looking at their expression for x^T(given above) it seems like they have said:

[tex]\left( {\left( {U^T } \right)^{ - 1} } \right)^T = \left( {\left( {U^T } \right)^T } \right)^{ - 1} = U^{ - 1} [/tex]

ie.They appear to have ignored the inverse symbol of (U^T)^-1. Considered U^T, taken the transpose to get U and then 'remembered' the inverse symbol again to produce (U^T)^-1 = U^-1. I'm unsure if this is a valid step. I've looked through the matrix identities I've got but I can't find one which suggests that what they've done to get the transpose of x is valid.
 
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  • #4
Benny said:
it seems like they have said:

[tex]\left( {\left( {U^T } \right)^{ - 1} } \right)^T = \left( {\left( {U^T } \right)^T } \right)^{ - 1} = U^{ - 1} [/tex]
ie.They appear to have ignored the inverse symbol of (U^T)^-1. Considered U^T, taken the transpose to get U and then 'remembered' the inverse symbol again to produce (U^T)^-1 = U^-1. I'm unsure if this is a valid step.

This looks ok to me because of the identities (for invertible nxn matrices):

[tex]\left(A^t\right)^{-1}=\left(A^{-1}\right)^t[/tex]

[tex]\left(A^t)\right^t=A[/tex]
 
  • #5
Oh ok I wasn't aware of the first identity.

I still don't understand how they came up with the substitution for x in the part (b) integral. If x is interpeted as a column vector then the argument of the exponential is a polynomial in x_i.

If a new 'variable' y where y is a column vector with entries y_i is introduced then exp((y^T)y) is equivalent to (exp(-z^2))^n where I have used z to avoid ambiguity. There would also be other parts to the integral but that's the main part of it. I just can't figure out a way to come up with a suitable substitution. I'm thinking that perhaps the integrand needs to be analysed but I just don't know where to start.
 
  • #6
Benny said:
Oh ok I wasn't aware of the first identity.
I still don't understand how they came up with the substitution for x in the part (b) integral. If x is interpeted as a column vector then the argument of the exponential is a polynomial in x_i.
If a new 'variable' y where y is a column vector with entries y_i is introduced then exp((y^T)y) is equivalent to (exp(-z^2))^n where I have used z to avoid ambiguity. There would also be other parts to the integral but that's the main part of it. I just can't figure out a way to come up with a suitable substitution. I'm thinking that perhaps the integrand needs to be analysed but I just don't know where to start.

Hey Benny. Sorry if I gave you the impression "I COULD" solve part 2. It's very interesting but I can't. Tell you what, if I had to first I would just solve it "directly" for n=2, then attempt to solve it "directly" for n=3, then maybe n=4, then look at the trend and then through induction attempt to come up with that nice matrix form of the solution.
 
  • #7
It's ok, the assignment questions I'll be doing will be different next year anyway.:biggrin: (This one is from a subject I didn't do this year)

I think it would be a good idea to consider trying it like you said, considering some specific cases. Anyway thanks for the help.

Edit: If I were to follow the suggestion in the solution, to use a procedure similar to the one for part then I guess the substitution for x can be gradually built up. Use the definition of A and hopefully the most convenient looking substitution will work.
 
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What is a definite integral with exponential?

A definite integral with exponential is a mathematical concept that involves finding the area under a curve where the function is defined by an exponential expression. It is used to solve problems related to growth and decay, such as compound interest or population growth.

What is the formula for calculating a definite integral with exponential?

The formula for calculating a definite integral with exponential is ∫eax dx = (1/a)eax + C, where a is the constant in the exponential expression and C is the constant of integration.

What is the difference between a definite integral and an indefinite integral?

A definite integral has specific limits of integration, while an indefinite integral does not. This means that a definite integral will give a specific numerical value, while an indefinite integral will give a general expression that includes a constant of integration.

What are some real-life applications of definite integrals with exponential?

Definite integrals with exponential are used in various fields, such as finance, economics, and biology. They can be used to calculate compound interest, population growth, and radioactive decay rates, among other things.

How do I solve a definite integral with exponential?

To solve a definite integral with exponential, you can use the formula mentioned earlier or use integration techniques such as substitution or integration by parts. It is important to correctly identify the limits of integration and use proper algebraic manipulation to solve the integral.

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