1. ## More improper integrals

Evaluate

$\displaystyle \int_0^1\frac{x^{2007}-x^{2006}}{\ln x}\,dx$

$\displaystyle \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}$

$\displaystyle \int_0^\infty\frac{dx}{(r^2+2rx\cos\theta+x^2)^{3/2}}$

2. 1)
Let $\displaystyle t=\ln x$ then:
$\displaystyle \int_{-\infty}^0 \frac{e^{2008t} - e^{2007t}}{t}dt$.
Thus,
$\displaystyle \int_{-\infty}^0 \int_0^{\infty} (e^{2008t}-e^{2007t})e^{-\mu t}~d\mu ~dt$

3. Originally Posted by liyi
Evaluate

$\displaystyle \int_0^1\frac{x^{2007}-x^{2006}}{\ln x}\,dx$
Hehe, I've got another solution for this one.

\displaystyle \begin{aligned} \int_0^1\frac{x^{2007}-x^{2006}}{\ln x}\,dx&=\int_0^1\int_{2006}^{2007}x^u\,du\,dx\\ &=\int_{2006}^{2007}\int_0^1x^u\,dx\,du\\ &=\int_{2006}^{2007}\frac{du}{u+1}\\ &=\ln\frac{2008}{2007}. \end{aligned}

You can also set $\displaystyle u=-\ln x$ and apply Frullani from there.

4. Originally Posted by liyi
$\displaystyle \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}$
$\displaystyle \frac{1}{\sqrt{(x-a)(b-x)}} = \frac{1}{\sqrt{-(x-a)(x-b)}}$.
Let $\displaystyle t=x - \frac{a+b}{2}$ and we get:
$\displaystyle \frac{1}{\sqrt{-\left( t - \frac{a-b}{2} \right)\left( t + \frac{a-b}{2} \right)}} = \frac{1}{\sqrt{\left( \frac{a-b}{2}\right)^2 - t^2}}$.
Thus the integral becomes,
$\displaystyle \int_{a-b}^{b-a} \frac{dt}{\sqrt{\left( \frac{a-b}{2}\right)^2 - t^2}} = \left. \frac{a-b}{2} \cdot \sin^{-1} \frac{2t}{a-b} \right|_{a-b}^{a+b}$

5. Another attempt.

Originally Posted by liyi
Evaluate

$\displaystyle \int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}$
Define $\displaystyle u=x-a,$

$\displaystyle \int_a^b {\frac{{dx}} {{\sqrt {(x - a)(b - x)} }}} = \int_0^{b - a} {\frac{{du}} {{\sqrt {u(b - u - a)} }}} .$

One more substitution according to $\displaystyle u = \frac{{(b - a)(1 + \kappa )}} {2},$

$\displaystyle \int_a^b {\frac{{dx}} {{\sqrt {(x - a)(b - x)} }}} = \int_{ - 1}^1 {\frac{{d\kappa }} {{\sqrt {1 - \kappa ^2 } }}} = 2\arcsin 1 = \pi .$

6. Originally Posted by liyi
Evaluate

$\displaystyle \int_0^\infty\frac{dx}{(r^2+2rx\cos\theta+x^2)^{3/2}}$
Rewrite the integral as

$\displaystyle \int_0^\infty {\frac{{dx}} {{\Big[ {\left( {x + r\cos \theta } \right)^2 + r^2 \sin ^2 \theta } \Big]\sqrt {\left( {x + r\cos \theta } \right)^2 + r^2 \sin ^2 \theta } }}} .$

Now here's the trick, make the substitution $\displaystyle u = \frac{{x + r\cos \theta }} {{r\sin \theta }}.$