# Definite integral

1. Jan 24, 2006

### ari_a

Can anyone help me integrating the folowwing integrand from zero to infinity:
x*ln^2(ax)*exp(-bx^2+cx)
where a,b and c are real and pisitive constants.
Thanks

Last edited: Jan 24, 2006
2. Jan 24, 2006

### maverick6664

First term diverges with $$x \rightarrow \infty$$. Is the region really from 0 to infinity? If possible, please post in TeX.
But first term is not difficult. Using partial integral,

$$\int x (\ln ax)^2 dx = 1/2 x^2 (\ln ax)^2 - \int \frac {ax^2} {ax} \ln ax dx = \frac 1 2 x^2 (\ln ax)^2 - \int x \ln ax dx$$.

Using partial integral on the second term, it is

$$\frac 1 2 x^2 \ln ax - \frac 1 2 \int ax^2 \frac 1 {ax} dx = \frac 1 2 x^2 \ln ax - \frac 1 4 x^2$$

So in total,

$$\int x (\ln ax)^2 dx = \frac 1 2 x^2((\ln ax))^2 - \ln ax) + \frac 1 4 x^2$$.

The second term can be integrated from $$-\infty \rightarrow \infty$$

Last edited: Jan 24, 2006
3. Jan 24, 2006

### maverick6664

As for $$\exp(-bx^2+cx)$$, it can be integrated for some kinds of limited regions...(AFAIK) using Gaussian integral. i.e. Put

$$I = \int_{-\infty}^\infty \exp(-x^2) dx$$.

Then

$$I^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty \exp(-(x^2 + y^2)) dx dy = 2 \pi \int_0^\infty r \exp(-r^2) dr = \pi$$

So $$I = \sqrt \pi.$$

But exp(-bx^2+cx) is another thing if the range is different.... If the region is from $$-\infty \rightarrow \infty$$

$$\int_{-\infty}^\infty \exp(-bx^2 + cx) dx = \int_{\infty}^\infty \exp(-b(x-\frac c {2b}) ^ 2 + \frac {c^2} {4b}) dx = \sqrt \pi \exp (c^2/4b) / \sqrt b$$

I cannot go further..

Last edited: Jan 24, 2006
4. Jan 24, 2006

### ari_a

I'm very sorry. I was mistaken. It should be a multiplication of the two terms, not sum. The integrand is now corrected in the original message.
Thanks