1. ## Two weird integrals

Compute integral (1 to 2) dx/((x)*(x-m)) if m is a small positive number.

What happens as m-->0+

Compute integral (1 to 2) dx/((x^2)+n) if n is a small positive number.

What happens when n-->0+

From the look of it, these integrals will be equal as both both n and m approach 0...the problem is I don't know how to show this.....

2. Originally Posted by zhupolongjoe
Compute integral (1 to 2) dx/((x)*(x-m)) if m is a small positive number.
$\int_1^2 \frac{dx}{x(x-m)}$
Use "partial fractions" to write $\frac{1}{x(x-m)}$ as $\frac{A}{x}+ \frac{B}{x- m}$. Then integrate (you will get logarithms).

What happens as m-->0+

Compute integral (1 to 2) dx/((x^2)+n) if n is a small positive number.
$\int_1^2 \frac{dx}{x^2+ n}$
Now that looks like an "arctan" problem to me. $\int \frac{dx}{x^2+ 1}= arctan(x)+ C$. If you let $\sqrt{n}u= x$ then $x^2+ n= nu^2+n= n(u^2+ 1)$. Of course, $dx= \sqrt{n} du$. When x= 1, u= $1/\sqrt{n}$ and when x= 2, u= $2/\sqrt{n}$ so your integral becomes [tex]\int_{1/\sqrt{n}}^{2/\sqrt{n}}\frac{\sqrt{n}du}{nu^2+ n}= 1/\sqrt{n}int_{1/\sqrt{n}}^{2/\sqrt{n}}\frac{du}{u^2+ 1}

What happens when n-->0+

From the look of it, these integrals will be equal as both both n and m approach 0...the problem is I don't know how to show this.....