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Math Help - Two weird integrals

  1. #1
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    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.....
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    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.....
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