# Definite integral with exponential

1. Dec 11, 2005

### Benny

Hi I have a few questions regarding the following two part problem.

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

$$Ans:e^{\frac{{b^2 }}{{4a}}} \sqrt {\frac{\pi }{a}}$$

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

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

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

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

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 $$x = \frac{1}{{\sqrt a }}\left( {y - c} \right)$$ (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 $$x = U^{ - 1} \left( {y + \frac{1}{2}U^{ - 1} b} \right)$$. I don't understand where they got it from. Could someone suggest how they would start this question? Any help would be good thanks.

2. Dec 11, 2005

### saltydog

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

$$e^{-a(x+k)^2}$$

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

$$\sigma=\sqrt{a}x$$

. . . thanks Tide.

3. Dec 11, 2005

### Benny

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: $$x = U^{ - 1} \left( {y + \frac{1}{2}\left( {U^T } \right)^{ - 1} b} \right)$$ where x, U, b etc are as defined in my initial post.

$$x^T = \left( {y^T + \frac{1}{2}b^T U^{ - 1} } \right)\left( {U^T } \right)^{ - 1}$$

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:

$$\left( {\left( {U^T } \right)^{ - 1} } \right)^T = \left( {\left( {U^T } \right)^T } \right)^{ - 1} = U^{ - 1}$$

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.

Last edited: Dec 11, 2005
4. Dec 11, 2005

### saltydog

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

$$\left(A^t\right)^{-1}=\left(A^{-1}\right)^t$$

$$\left(A^t)\right^t=A$$

5. Dec 11, 2005

### Benny

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. Dec 12, 2005

### saltydog

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. Dec 12, 2005

### Benny

It's ok, the assignment questions I'll be doing will be different next year anyway. (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.

Last edited: Dec 12, 2005