MHB Is McLaurin Expansion the Key to Solving Integration by Expansion?

AI Thread Summary
The discussion focuses on evaluating the integral I(x) = (1/π) ∫[0,π] sin(xsint) dt and demonstrating that I(x) = (2x/π) + O(x^3) as x approaches 0. The McLaurin series expansion for sin(xsint) is introduced, leading to the uniform convergence of the series on the interval [0, π]. This uniform convergence allows for the interchange of summation and integration, simplifying the evaluation of the integral. The conclusion suggests that the remaining steps can be completed based on this established framework.
ra_forever8
Messages
106
Reaction score
0
Consider the integral
\begin{equation}
I(x)= \frac{1}{\pi} \int^{\pi}_{0} sin(xsint) dt
\end{equation}
show that
\begin{equation}
I(x)= \frac{2x}{\pi} +O(x^{3})
\end{equation}
as $x\rightarrow0$.
=> I Have used the expansion of McLaurin series of $I(x)$ but did not work.
please help me.
 
Last edited:
Mathematics news on Phys.org
I'm not sure if the expansion correct. However, we can do it as follow.

By McLaurin expansion, we have

$$f(x,t)=\sin(x\sin t)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}(\sin t)^{2n-1}=\sum_{n=1}^{\infty} f_n(x,t)$$.

Since we integrate $$f(x,t)$$ on $$E=[0,\pi]\ni t$$, we should test uniform convergence of the series, $$\sum_{n=1}^{\infty} f_n(x,t)$$, at first.

Clearly, $$|f_n(x,t)|\leq \frac{|x|^{2n-1}}{(2n-1)!}$$ on $$E\ni t$$, and $$\sum_{n=1}^{\infty} \frac{|x|^{2n-1}}{(2n-1)!}$$ converges uniformly on $$\mathbb{R}\ni x$$. Thus, $$\sum_{n=1}^{\infty} f_n(x,t)$$ converges uniformly on $$E\ni t$$.Hence $$\frac{1}{\pi}\int_E f(x,t)dt=\frac{1}{\pi}\int_E \sum_{n=1}^{\infty}f_n(x,t)dt=\frac{1}{\pi}\sum_{n=1}^{\infty}\int_E f_n(x,t)dt=\frac{1}{\pi}\int_E x\sin tdt+O(x^3)$$, as $$x$$ goes to $$0$$.

I think you can complete the rest.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Back
Top