Three impossible integrals and two challenging ones

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May 18th 2008, 08:58 PM
Mathstud28

Three impossible integrals and two challenging ones

Show that

$\int_0^{\frac{\pi}{2}}\tan^{x}(\theta)d\theta=\fra c{\pi}{2\cos\bigg(\frac{x\pi}{2}\bigg)}$

and

$\int_0^{\infty}\frac{arctan(ax)-arctan(bx)}{x}=\frac{\pi}{2}\ln\bigg(\frac{a}{n}\b igg)$

$\int_0^{\frac{\pi}{2}}\sqrt{\cos(x)}dx=\frac{(2\pi )^{\frac{3}{2}}}{[\Gamma(\frac{1}{4})]^2}$

and

$\int_0^{\frac{\pi}{2}}\frac{dx}{(a^2\sin^2(x)+b^2\ cos^2(x))^2}=\frac{\pi(a^2+b^2)}{4a^3b^3}$

and finally

Show that the area enclosed by a curve defined through the equation

$x^{\frac{b}{c}}+y^{\frac{b}{c}}=a^{\frac{b}{c}}$

where $a>0$

c is a positive odd integer and b is a positive even integer is

$\frac{\bigg[\Gamma\bigg(\frac{c}{b}\bigg)\bigg]^2}{\Gamma\bigg(\frac{2c}{b}\bigg)}\cdot\bigg(\fra c{2ca^2}{b}\bigg)$

Come on people lets see that skill (Sun)
May 18th 2008, 10:02 PM
NonCommAlg

well, they are actually pretty easy. i'll do the first three for now:

Quote:

Originally Posted by Mathstud28

Show that

$\int_0^{\frac{\pi}{2}}\tan^{x}(\theta)d\theta=\fra c{\pi}{2\cos\bigg(\frac{x\pi}{2}\bigg)}$

here we need to have $|x| < 1.$ then we have:

$\int_0^{\frac{\pi}{2}} \tan^{x} \theta \ d\theta = \int_0^{\frac{\pi}{2}} \sin^x \theta \cos ^{-x} \theta \ d\theta = \frac{1}{2}B \left(\frac{1+x}{2}, \frac{1-x}{2} \right)$

$=\frac{1}{2}\Gamma \left(\frac{1+x}{2} \right) \Gamma \left(1 - \frac{1+x}{2} \right)=\frac{\pi}{2\sin(\frac{1+x}{2} \pi)}$ by Euler's Reflection formula

$=\frac{\pi}{2 \cos(\frac{\pi x}{2})}. \ \ \square$

Quote:

$\int_0^{\infty}\frac{arctan(ax)-arctan(bx)}{x}=\frac{\pi}{2}\ln\bigg(\frac{a}{n}\b igg)$

that $n$ is actually $b.$ i'll assume 0 < b < a. we have:

$\int_0^{\infty} \frac{\arctan(ax) - \arctan(bx)}{x} dx= \int_0^{\infty} \int_b^a \frac{1}{1+x^2y^2}dy dx$

$=\int_b^a \int_0^{\infty} \frac{1}{1 + x^2y^2} dxdy=\int_b^a \frac{\pi}{2y}dy=\frac{\pi}{2}\ln(\frac{a}{b}). \ \ \ \square$

Quote:

$\int_0^{\frac{\pi}{2}}\sqrt{\cos(x)} dx=\frac{(2\pi)^{\frac{3}{2}}}{[\Gamma(\frac{1}{4})]^2}$

$\int_0^{\frac{\pi}{2}} \sqrt{\cos x} \ dx =\frac{1}{2}B \left(\frac{1}{2},\frac{3}{4}\right)=\frac{\Gamma( \frac{1}{2}) \Gamma(\frac{3}{4})}{2\Gamma(\frac{5}{4})}=\frac{\ sqrt{\pi} \Gamma(\frac{1}{4}) \Gamma(\frac{3}{4})}{2\Gamma(\frac{5}{4})\Gamma(\f rac{1}{4})}. \ \ \ \ \ (*)$

now, again, by Euler's Reflection formula: $\Gamma(\frac{1}{4}) \Gamma(\frac{3}{4})=\frac{\pi}{\sin(\frac{\pi}{4}) }=\pi \sqrt{2},$ and

clearly $\Gamma(\frac{5}{4})=\frac{1}{4} \Gamma(\frac{1}{4}).$ put these two in $(*)$ to get the result. Q.E.D.
May 19th 2008, 06:09 AM
Krizalid

Quote:

Originally Posted by Mathstud28

Show that

$\int_0^{\infty}\frac{arctan(ax)-arctan(bx)}{x}=\frac{\pi}{2}\ln\bigg(\frac{a}{n}\b igg)$

See here.