Series expansion of Sine Integral Si(x)

[yungman]

This is not really a homework, I am trying to expand Si(x) into a series.The series expansion of Si(x) is given in articles:
$$Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^{\infty}\frac {(-1)^k x^{2k+1}}{(2k+1)(2k+1)!}$$

This is my work, I just cannot get the right answer:
$$Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^m \frac{sin \theta_i}{\theta_i}\Delta\theta_i$$

$$\sin \theta=\sum_0^{\infty}\frac{(-1)^k \theta^{2k+1}}{(2k+1)!}\Rightarrow\;\frac{\sin\theta}{\theta}=\sum_0^{\infty}\frac{(-1)^k \theta^{2k}}{(2k+1)!}$$

$$\Delta \theta_i = \frac x m\;\hbox{ and }\; \theta_i= i\Delta \theta_i =i\frac x m$$

$$\Rightarrow\;Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^m \frac{sin \theta_i}{\theta_i}\Delta\theta_i\;=\;\sum_0^m\sum_0^{\infty}\frac{(-1)^k\left (i\frac x m\right )^{2k}}{(2k+1)!}\frac x m\;=\;\sum_0^m\sum_0^{\infty}\frac{(-1)^ki^{2k} x^{2k+1}}{m^{2k+1}(2k+1)!}$$

I don't know how to get the answer, please help.

 

[Dick]

This is not really a homework, I am trying to expand Si(x) into a series.The series expansion of Si(x) is given in articles:
$$Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^{\infty}\frac {(-1)^k x^{2k+1}}{(2k+1)(2k+1)!}$$

This is my work, I just cannot get the right answer:
$$Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^m \frac{sin \theta_i}{\theta_i}\Delta\theta$$

$$\sin \theta=\sum_0^{\infty}\frac{(-1)^k \theta^{2k+1}}{(2k+1)!}\Rightarrow\;\frac{\sin\theta}{\theta}=\sum_0^{\infty}\frac{(-1)^k \theta^{2k}}{(2k+1)!}$$

$$\Delta \theta_i = \frac x m\;\hbox{ and }\; \theta_i= i\Delta \theta_i =i\frac x m$$

$$\Rightarrow\;Si(x)=\int_0^x \frac{\sin\theta}{\theta}d\theta=\sum_0^m \frac{sin \theta_i}{\theta_i}\Delta\theta_i\;=\;\sum_0^m\sum_0^{\infty}\frac{(-1)^k\left (i\frac x m\right )^{2k}}{(2k+1)!}\frac x m\;=\;\sum_0^m\sum_0^{\infty}\frac{(-1)^ki^{2k} x^{2k+1}}{m^{2k+1}(2k+1)!}$$

I don't know how to get the answer, please help.

I have no idea what all the delta thetas are all about. You don't need to derive the formula for integrating a power. You have a series for $\sin(\theta)$, divide it by $\theta$ and integrate term by term.

 

[yungman]

I have no idea what all the delta thetas are all about. You don't need to derive the formula for integrating a power. You have a series for $\sin(\theta)$, divide it by $\theta$ and integrate term by term.

I missed the subscript i, $\Delta\theta_i$ is just the series representation of $d\theta$. Where I chop the 0<$\theta$<x into m sections.

I want to verify and derive the equation given from articles.

 

[Dick]

$\Delta\theta$ is just the series representation of $d\theta$. Where I chop the 0<$\theta$<x into m sections.

I want to verify and derive the equation given from articles.

You don't have to chop into sections. Just integrate the power series term by term!

 

[yungman]

You don't have to chop into sections. Just integrate the power series term by term!

I got it! It's just that easy! Thanks.