Annoying integral

**Mathmo55** · April 15th 2010, 07:22 AM

Let $f(x) = \log{(1-x^2})$

Let $I_n = \int_{n-\frac{1}{2}}^{n+\frac{1}{2}} \log{x} \ dx - \log{n}$ for n natural.

Show that $I_n = \int_0^{\frac{1}{2}} f\left(\frac{t}{n}\right) dt$

I just can't get this. Any help would be appreciated. Thanks

**simplependulum** · April 15th 2010, 07:35 AM

Originally Posted by Mathmo55

Let $f(x) = \log{(1-x^2})$

Let $I_n = \int_{n-\frac{1}{2}}^{n+\frac{1}{2}} \log{x} \ dx - \log{n}$ for n natural.

Show that $I_n = \int_0^{\frac{1}{2}} f\left(\frac{t}{n}\right) dt$

I just can't get this. Any help would be appreciated. Thanks

$\int_{n-\frac{1}{2}}^{n+\frac{1}{2}} = \int_{n-\frac{1}{2}}^n + \int_n^{n+\frac{1}{2}}$

For the first one , substitute $x = n-t$

For the second one , substitute $x = n+t$

I think we will see such thing :

$I_n = \int_0^{\frac{1}{2}} [\log(n-t) + \log(n+t)]~dt -\log{n}$

$= \int_0^{\frac{1}{2}} \log(n^2-t^2) ~dt -\log{n}$

$= \int_0^{\frac{1}{2}} [\log(n^2) + f(\frac{t}{n}) ]~dt -\log{n}$

$= \log{n} + \int_0^{\frac{1}{2}} f(\frac{t}{n})~dt - \log{n}$

$= \int_0^{\frac{1}{2}} f(\frac{t}{n})~dt$

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