Re: Trigonometric times Exponential
Posted: Tue Jul 12, 2016 6:59 am
by Papapetros Vaggelis
Hello Tolis. I give a solution but i don't know if it coincides with yours. If not, post yours.
As i can imagine, your solution uses methods of complex analysis.
First of all, I''ll calculate the integrals
\(\displaystyle{I:\mathbb{N}\times \mathbb{N}\longrightarrow \mathbb{R}\,,I(n,m)=\int_{0}^{\frac{\pi}{2}}e^{n\,x}\,\cos\,(m\,x)\,\mathrm{d}x}\) .
\(\displaystyle{\begin{aligned}I(n,m)&=\int_{0}^{\frac{\pi}{2}}e^{n\,x}\,\cos\,(m\,x)\,\mathrm{d}x\\&=\left[\dfrac{e^{n\,x}}{n}\,\cos\,(m\,x) \right ]_{0}^{\frac{\pi}{2}}+\dfrac{m}{n}\,\int_{0}^{\frac{\pi}{2}}e^{n\,x}\,\sin\,(m\,x)\,\mathrm{d}x\\&=\dfrac{e^{n\,\pi/2}}{n}\,\cos\,(m\,\pi/2)-\dfrac{1}{n}+\dfrac{m}{n}\,\left[\left[\dfrac{e^{n\,x}}{n}\,\sin\,(m\,x) \right ]_{0}^{\frac{\pi}{2}}-\dfrac{m}{n}\,\int_{0}^{\frac{\pi}{2}}e^{n\,x}\,\cos\,(m\,x) \right ]\\&=\dfrac{e^{n\,\pi/2}}{n}\,\cos\,(m\,\pi/2)-\dfrac{1}{n}+\dfrac{m}{n^2}\,e^{n\,\pi/2}\,\sin\,(m\,\pi/2)-\dfrac{m^2}{n^2}\,I(n,m) \end{aligned}}\)
so :
\(\displaystyle{\left(\dfrac{n^2+m^2}{n^2}\right)\,I(n,m)=\dfrac{n\,e^{n\,\pi/2}\,\cos\,(m\,\pi/2)-n+m\,e^{n\,\pi/2}\,\sin\,(m\,\pi/2)}{n^2}}\) ,
or :
\(\displaystyle{I(n,m)=\dfrac{n\,e^{n\,\pi/2}\,\cos\,(m\,\pi/2)-n+m\,e^{n\,\pi/2}\,\sin\,(m\,\pi/2)}{n^2+m^2}\,,n\,,m\in\mathbb{N}}\) .
It's known that \(\displaystyle{\cos\,(3\,x)=4\,\cos^3\,x-3\,\cos\,x\,,x\in\mathbb{R}}\)
and
\(\displaystyle{\begin{aligned} \sin^4\,x&=\left(\sin^2\,x\right)^2\\&=\left(\dfrac{1}{2}-\dfrac{\cos\,2\,x}{2}\right)^2\\&=\dfrac{1}{4}-\dfrac{1}{2}\,\cos\,(2\,x)+\dfrac{1}{4}\,\cos^2\,2\,x\\&=\dfrac{1}{4}-\dfrac{1}{2}\,\cos\,(2\,x)+\dfrac{1+\cos\,4\,x}{8}\\&=\dfrac{1}{8}\,\cos\,4\,x-\dfrac{1}{2}\,\cos\,2\,x+\dfrac{3}{8}\,,x\in\mathbb{R}\end{aligned}}\)
Therefore :
\(\displaystyle{\begin{aligned} \int_{0}^{\frac{\pi}{2}}e^{3\,x}\,\cos^3\,x\,\mathrm{d}x&=\int_{0}^{\frac{\pi}{2}}e^{3\,x}\,\left[\dfrac{1}{4}\,\cos\,(3\,x)+\dfrac{3}{4}\,\cos\,x\right]\,\mathrm{d}x\\&=\dfrac{1}{4}\,\int_{0}^{\frac{\pi}{2}}e^{3\,x}\,\cos\,(3\,x)\,\mathrm{d}x+\dfrac{3}{4}\,\int_{0}^{\frac{\pi}{2}}e^{3\,x}\,\cos\,x\,\mathrm{d}x\\&=\dfrac{1}{4}\,I(3,3)+\dfrac{3}{4}\,I(3,1)\end{aligned}}\)
and :
\(\displaystyle{\begin{aligned} \int_{0}^{\frac{\pi}{2}}e^{4\,x}\,\sin^4\,x\,\mathrm{d}x&=\int_{0}^{\frac{\pi}{2}}e^{4\,x}\,\left[\dfrac{3}{8}\,+\dfrac{1}{8}\,\cos\,(4\,x)-\dfrac{1}{2}\,\cos\,(2\,x)\right]\,\mathrm{d}x\\&=\dfrac{3}{8}\,\int_{0}^{\frac{\pi}{2}}e^{4\,x}\,\mathrm{d}x+\dfrac{1}{8}\,\int_{0}^{\frac{\pi}{2}}e^{4\,x}\,\cos\,(4\,x)\,\mathrm{d}x-\dfrac{1}{2}\,\int_{0}^{\dfrac{\pi}{2}}e^{4\,x}\,\cos\,(2\,x)\,\mathrm{d}x\\&=\left[\dfrac{3\,e^{4\,x}}{32}\right]_{0}^{\frac{\pi}{2}}+\dfrac{1}{8}\,I(4,4)-\dfrac{1}{2}\,I(4,2)\\&=\dfrac{3}{32}\,e^{2\,\pi}-\dfrac{3}{32}+\dfrac{1}{8}\,I(4,4)-\dfrac{1}{2}\,I(4,2)\end{aligned}}\)