# Evaluating sum $\sum_{m=0}^{\infty}\frac{(2-\delta_m^0)(-1)^m \lambda_0}{a(\lambda_0^2 -(\frac{m\pi}{a}))}\cos(m\pi x/a)$

How Can I evaluate the following sum$$\sum_{m=0}^{\infty}\frac{2-\delta_m^0}{a}\frac{(-1)^m \lambda_0}{\lambda_0^2 -(\frac{m\pi}{a})}\cos\left(\frac{m\pi x}{a}\right)=\frac{\cos(\lambda_0 x)}{\sin(\lambda_0 a)}$$

I have read this in a research paper I have tried evaluating the sum using finite cosine transform We have $$\frac{2}{a}\frac{1}{\lambda_0}+\frac{2}{a}\sum_{m=0}^{\infty}\frac{(-1)^m \lambda_0}{\lambda_0^2 -(\frac{m\pi}{a})}\cos\left(\frac{m\pi x}{a}\right)=f(x)$$ So $$\frac{(-1)^m \lambda_0}{\lambda_0^2 -(\frac{m\pi}{a})}=\int_{0}^{a} f(x)\cos\left(\frac{m\pi x}{a}\right) dx$$

How to find $$f(x)$$ ?

And Is there any other way to evaluate the sum ?

• $\displaystyle 1/\lambda_{0}$, $\displaystyle x$ and $\displaystyle a$ seem to have the same dimensions. So, something is wrong with the original statement. – Felix Marin May 18 '19 at 5:39
• How is it that your LHS is $2a$ periodic but your RHS isn't? Otherwise, this would just be the Fourier series of your function in the interval $\left[-a,a\right]$. – eranreches May 18 '19 at 16:15
• what do you meanby $\delta_m^0$? – G Cab May 19 '19 at 23:01
• $\delta^0_m =1$if m=0 and =0 otherwise – Mahmoud Hassan May 20 '19 at 18:36
• Can you specify the domain of the constants $a$ and $\lambda_0$? Integer, real, or complex? Is $a>0$? I'm not sure your formula for $f(x)$ is correct. Note $\sum_{m=0}^\infty\frac{(2-\delta_{0,m})\,(-1)^m\,\lambda_0}{a \left(\lambda_0^2-\frac{m \pi}{a}\right)} \cos \left(\frac{m \pi x}{a}\right)=\frac{1}{a \lambda_0}+\frac{2 \lambda_0}{a}\sum_{m=1}^{\infty } \frac{(-1)^m }{\lambda_0^2-\frac{m \pi}{a}}\cos \left(\frac{m \pi x}{a}\right)$. – Steven Clark May 20 '19 at 22:20

Focusing on the main task of the problem in the OP I shall calculate the sum starting at $$m=1$$ and dropping the overall factor $$\lambda$$, i.e. the expression

$$f(x) = \frac{2 }{a} \sum _{m=1}^{\infty } \frac{(-1)^m \cos \left(\frac{\pi m x}{a}\right)}{\lambda ^2-\frac{\pi m}{a}}\tag{1}$$

Assuming $$\lambda ^2 \lt \frac{\pi}{a}$$ we write

$$\frac{1}{\lambda ^2-\frac{\pi m}{a}} = - \int_{0}^{\infty} e^{-t (\frac{\pi m}{a}-\lambda ^2)}\, dt\tag{2}$$

and

$$\cos(\frac{m \pi x}{a}) = \Re \exp(i \frac{m \pi x}{a})$$

Substituting this into $$(1)$$ the /geometric) sum can be done under the integral:

$$\sum _{m=1}^{\infty } (-1)^m \exp \left(\frac{i \pi m x}{a}\right) \exp \left(-t \left(\frac{\pi m}{a}-\lambda ^2\right)\right)\\= -\frac{e^{t \left(-\left(\frac{\pi }{a}-\lambda ^2\right)\right)+\frac{\pi t}{a}+\frac{i \pi x}{a}}}{e^{\frac{\pi t}{a}}+e^{\frac{i \pi x}{a}}}$$

Integrating the negative this over $$t$$ according to $$(2)$$ and applying the missing factor gives

$$-\frac{2}{\pi} \left(-e^{\frac{i \pi x}{a}}\right)^{\frac{a \lambda ^2}{\pi }} B(-e^{\frac{i \pi x}{a}},1-\frac{a \lambda ^2}{\pi },0)\tag{3}$$

Here $$B$$ is the incomplete Beta function defined by

$$B(z,a,b) = \int_{0}^{z} t^{a-1} (1-t)^{b-1}$$

We now have to take the real part.

Selecting the simplified case $$\lambda \to 0$$ expression $$(3)$$ reduces to

$$\frac{2}{\pi} \log \left(1+e^{\frac{i \pi x}{a}}\right)$$

The real part and hence the sum is

$$f(\lambda\to 0)=\frac{1}{\pi} \log \left(4 \cos \left(\frac{\pi x}{2 a}\right)^2\right)\tag{4}$$

The case $$0 \lt \lambda ^2 (\lt \frac{\pi}{a})$$ can be extracted from $$(3)$$ as well. It will be left as an exercise in complex arithmetic to the reader.

For any $$\lambda \ne \sqrt{\frac{\pi m}{a}}$$ the sum can be expressed by the Hurwitz-Lerch transcendent as follows

$$f= \frac{1}{\pi} \left( e^{\frac{i \pi x}{a}} \Phi \left(-e^{\frac{i \pi x}{a}},1,1-\frac{a \lambda ^2}{\pi }\right) +e^{-\frac{i \pi x}{a}} \Phi \left(-e^{-\frac{i \pi x}{a}},1,1-\frac{a \lambda ^2}{\pi }\right) \right)\tag{5}$$

$$\Phi$$ is defined as

$$\Phi(z,s,a) = \sum_{k=0}^{\infty} z^k (k+a)^{-s}$$

• Thank you very much . – Mahmoud Hassan May 23 '19 at 11:08
• I'll try to follow up the calculation for $(m \pi /a)^2$ to get $\frac{\cos(\lambda x)}{\sin(\lambda a)}$ – Mahmoud Hassan May 23 '19 at 11:32
• @ Mahmoud Hassan Hint: after partial fraction decomposition you can use my results immediately. However, you won't get $\frac{\cos(\lambda x)}{\sin(\lambda x)}$. – Dr. Wolfgang Hintze May 24 '19 at 7:24

Too long for the comment.

The issue sum $$S_1 = \sum_{m=0}^{\infty}\frac{2-\delta_{m0}}{a}\frac{(-1)^m \lambda_0}{\lambda_0^2 -(\frac{m\pi}{a})}\cos\left(\frac{m\pi x}{a}\right),\tag1$$

where $$\delta_{ij}$$ is the Kronekker symbol,

can be considered as the particular case of $$S_k$$, where $$S_k = \frac{1}{a}\frac{1}{\lambda_0} + \frac{2}{a}\sum_{m=1}^{\infty}\frac{(-1)^m \lambda_0}{\lambda_0^2 -(\frac{m\pi}{a})^k}\cos\left(\frac{m\pi x}{a}\right).\tag2$$

Using known exponential Fourier series for cosine function

$$\cos(bz) = \dfrac{2b\sin(b\pi)}\pi\left(\dfrac1{2b^2}-\sum\limits_{m=1}^\infty \dfrac{(-1)^m\cos(mz)}{m^2-b^2}\right)\tag3$$

with the values $$b=\dfrac{\lambda_0a}\pi,\quad z = \dfrac{\pi x}a,$$

one can get $$\dfrac 1\pi\dfrac{\cos(\lambda_0x)}{\sin(\lambda_0a)} = \dfrac1{a\lambda_0} + \dfrac{2\lambda_0a}{\pi^2} \sum\limits_{m=1}^\infty \dfrac{(-1)^m\cos\left(\dfrac{m\pi x}{a}\right)}{\dfrac{\lambda_0^2a^2}{\pi^2}-m^2}=S_2.$$