1. ## Annoying integral

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

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

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

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

2. Originally Posted by Mathmo55
Let $\displaystyle f(x) = \log{(1-x^2})$

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

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

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

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

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

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

I think we will see such thing :

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

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

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

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

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