User:Egm6341.s10.team4.anandankala/HW7-15

=Integral Evaluation=

Given
$$\displaystyle f(cos\theta)= \frac{a_0}{2} + \sum_{k=1}^{\infty} a_kcos(k\theta)$$

$$\displaystyle I= \int_{-1}^{1}f(x)dx= \int_{0}^{\pi}f(cos\theta)sin\theta d\theta$$

Prove that
$$\displaystyle I= a_0 + \sum_{j=1}^{\infty} \frac{2a_{2j}}{1-(2j)^2}$$

Solution
$$\displaystyle I= \int_{-1}^{1}f(x)dx= \int_{0}^{\pi}f(cos\theta)sin\theta d\theta$$

$$\displaystyle \Rightarrow I= \int_{0}^{\pi}(\frac{a_0}{2}+ \sum_{k=1}^{\infty}a_kcos(k\theta))sin\theta d\theta$$

$$\displaystyle \Rightarrow I= \int_{0}^{\pi} \frac{a_0}{2} sin\theta d\theta + \sum_{k=1}^{\infty}\int_{0}^{\pi} a_kcos(k\theta))sin\theta d\theta $$

$$\displaystyle \Rightarrow I= \dfrac{a_0}{2} [cos\theta]_{\pi}^{0}+ \sum_{k=1}^{\infty}\int_{0}^{\pi} \dfrac{a_k}{2} [sin((1+k)\theta) +sin((1-k)\theta)]d\theta $$

$$\displaystyle \Rightarrow I= a_0+ \sum_{k=1}^{\infty}\dfrac{a_k}{2} [\int_{0}^{\pi} sin((1+k)\theta)d\theta+\int_{0}^{\pi} sin((1-k)\theta)d\theta] $$

$$\displaystyle \Rightarrow I= a_0+ \sum_{k=1}^{\infty} \dfrac{a_k}{2} [[\dfrac{cos((1+k)\theta)}{(1+k)}]_{\pi}^{0} +[\dfrac{cos((1-k)\theta)}{(1-k)}]_{\pi}^{0}]$$

Since all the cosine terms for odd values of k become zero. The above expression is modified as below.

$$\displaystyle \Rightarrow I= a_0+ \sum_{j=1}^{\infty} \dfrac{a_{2j}}{2} [[\dfrac{cos((1+2j)\theta)}{(1+2j)}]_{\pi}^{0} +[\dfrac{cos((1-2j)\theta)}{(1-2j)}]_{\pi}^{0}]$$

$$\displaystyle \Rightarrow I= a_0+ \sum_{j=1}^{\infty} \dfrac{a_{2j}}{2} \dfrac{2}{(1+2j)}]+[\dfrac{2}{(1-2j)}$$

$$\displaystyle \Rightarrow I= a_0+ \sum_{j=1}^{\infty} a_{2j}\dfrac{1}{(1+2j)}]+[\dfrac{1}{(1-2j)}$$

$$\displaystyle \Rightarrow I= a_0+ \sum_{j=1}^{\infty} \dfrac{2a_{2j}}{1-(2j)^2}.$$

Hence the proof.