0
1.1kviews
Show that $\int_{0}^{\infty}cos(ax^{\frac{1}{n}})dx=\frac{\Gamma(n+1)} {a^n}COS\frac{n\pi}{2}\cdot (H)$
1 Answer
0
0views

Put,

$ax^{1/n}=t \hspace{0.3cm}=\gt\hspace{0.3cm} x^{1/n}=t/a \hspace{0.3cm}=\gt x=(\frac{t}{a})^n $

$dx =n(\frac{t}{a})^{n-1}*\frac{1}{a}dt$

$\int_{0}^{\infty} cos\,t \frac{n}{a^n}t^{n-1}dt =\frac{n}{a^n}\int_{0}^{\infty}cos\,t \,t^{n-1}dt$

R.P

$\frac{n}{a^n}\int_{0}^{\infty}e^{-it}\,t^{n-1}dt$

Put,

$it=u \hspace{0.3cm}=\gt\hspace{0.3cm}idt=du $

R.P,

$\frac{n}{a^n} \int_{0}^{\infty} e^{-u} (\frac{u}{i})^{n-1} \frac{du}{i}=\frac{n}{a^n\,i^n} \int_{0}^{\infty} e^{-u} u^{n-1} du $

$\hspace{4.5cm}= \frac{n}{a^n\,i^n}\sqrt{n}$

R.P

$\frac{\sqrt{n}+1}{a^n} i^{-n}$

R.P

$\frac{\sqrt{n}+1}{a^n} (cos\,\frac{\pi}{2}+i\,sin\,\frac{\pi}{2})^{-n}$

R.P,

$\frac{\sqrt{n}+1}{a^n} (cos\,n\frac{\pi}{2}-i\,sin\,n\frac{\pi}{2})\hspace{0.3cm}$

(Demoivre's Theoram)

$\frac{\sqrt{n}+1}{a^n} \, cos\,n\frac{\pi}{2}$

Please log in to add an answer.