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Show that $$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}k \sin(ax)}{a^{2}+k^{2}}=\frac{\pi}{2}\frac{\sinh(ax)}{\sinh(\pi a)}, \;\ x\in (-\pi,\pi)$$

It appears to me this series is crying out for the use of Fourier series. It seems to me I get close, but am failing to put the pieces all together.

Thus, I tried using the Fourier series for $f(x)=e^{ax}$. $\qquad \cosh(ax)$ gives the same thing.

$$a_{k}=\frac{1}{\pi}\int_{-\pi}^{\pi}e^{ax} \cos(kx)dx=\frac{e^{a\pi}(a \cos(k\pi)+k \sin(k\pi))}{\pi (a^{2}+k^{2})}-\frac{e^{-a\pi}(a \cos(k\pi)-k \sin(k \pi))}{\pi (a^{2}+k^{2})}$$

$b_{k}=0$

The given series is evident amongst $a_{n}$, but I kind of get hung up.

$\displaystyle a_{0}=\frac{2 \sinh(\pi a)}{\pi a}$

Now, using $\displaystyle e^{ax}=\frac{a_{0}}{2}+\sum_{k=1}^{\infty}a_{k} \cos(kx)\Rightarrow e^{ax}=\frac{\sinh(\pi a)}{\pi a}+\sum_{k=1}^{\infty}a_{k} \cos(kx)$

I even tried equating real and imaginary parts. Where the definite integral becomes

$$\left(\frac{(a \cos(k\pi)+k \sin(k\pi))e^{a \pi}}{a^{2}+k^{2}}+\frac{(a \sin(k\pi)-k \cos(k\pi))e^{a\pi}}{a^{2}+k^{2}}i\right)$$ $$-\left(\frac{(a \cos(k\pi)-k \sin(k\pi))e^{-\pi a}}{a^{2}+k^{2}}-\frac{(a \sin(k\pi)+k \cos(k\pi))e^{-a\pi}}{a^{2}+k^{2}}i\right)$$

I think I can equate the imaginary and real parts:

$$e^{ax}=\frac{e^{\pi a}-e^{-\pi a}}{2}\sum_{k=1}^{\infty}..............$$

I get mixed up here. The identity in the previous line is $ \sinh(\pi a)$, so it looks like I am onto something. If that $e^{ax}$ on the left of the equals sign were $ \sinh(ax)$, then a little algebra and I would practically be done.

I tried to get to the known sum, but can't quite get there.

There is something I am overlooking. Can anyone point me in the right direction, please?.

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The right side has a $k$ but no $x$. I assume you mean $\sin(ak)/\sin(a\pi)$. Also of course you must assume $a$ is not an integer. And surely it's the Fourier series for $\sin(ax)$ that you're looking for. –  Robert Israel May 29 '11 at 23:43
    
Oops, I meant that you mean $\sin(ax)/\sin(a\pi)$. Hmm, looks to me like that should be $a^2 - k^2$, not $a^2 + k^2$, in the denominator. –  Robert Israel May 29 '11 at 23:49
    
Yes, sorry, I had a typo in the first line. It was supposed to be sinh as well instead of sin. I fixed it. It was given as $a^{2}+k^{2}$ in the denominator. I tried the series for sin(ax), but it is 0. $e^{ax}$ or $cosh(ax)$ appears to be the ones here because they are used in the fourier evaluation of the series $\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^{2}+1}$. This is what gave me the idea, but I got stuck. –  Cody May 30 '11 at 0:37
    
It should be $\sin(kx)$ on the left hand side instead of $\sin(ax)$ –  user17762 May 30 '11 at 4:32
    
+1 for the beuty! –  Arjang May 30 '11 at 5:44

1 Answer 1

up vote 12 down vote accepted

Consider $\sinh(ax)$. Note that it is an odd function and hence if we expand it in sine's and cosine's the coefficient of cosines will be zero.

Hence, we can write $\displaystyle \sinh(ax) = \sum_{k=1}^{\infty} a_k \sin(kx)$

$$a_k = \frac1{\pi} \int_{-\pi}^{\pi} \sinh(ax) \sin(kx) dx$$

$$I = \int_{-\pi}^{\pi} \sinh(ax) \sin(kx) dx = \int_{-\pi}^{\pi} \left(\frac{e^{ax} - e^{-ax}}{2} \right) \left( \frac{e^{ikx} - e^{-ikx}}{2i} \right) dx $$ $$I = \frac1{4i} \left( \frac{e^{(a+ik)x}}{a+ik} - \frac{e^{(a-ik)x}}{a-ik} - \frac{e^{(-a+ik)x}}{-a+ik} - \frac{e^{-(a+ik)x}}{a+ik} \right)_{-\pi}^{\pi}$$ $$I = \frac{2}{4i} \left( \frac{(-1)^k e^{a \pi}}{a+ik} - \frac{(-1)^k e^{a \pi}}{a-ik} - \frac{(-1)^k e^{-a \pi}}{-a+ik} - \frac{(-1)^k e^{-a \pi}}{a+ik} \right)$$ $$I = \frac{(-1)^k}{2i} \left( \frac{-2 i k e^{a \pi}}{a^2+k^2} + \frac{2 i k e^{-a \pi}}{a^2+k^2} \right) = (-1)^{k-1} \frac{2 k }{a^2 + k^2} \sinh(a \pi)$$ Hence, we have $$\sinh(ax) = \frac{ 2 \sinh(a \pi) }{\pi} \sum_{k=1}^{\infty} (-1)^{k-1} \frac{k}{a^2 + k^2} \sin(kx)$$ Rewriting, we get the desired result, namely $$\sum_{k=1}^{\infty} \frac{ (-1)^{k-1} k \sin(kx)}{a^2 + k^2} = \frac{\pi}{2} \frac{\sinh(ax)}{\sinh(a \pi)}$$

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+1 for the nice answer! –  Arjang May 30 '11 at 5:45
    
Thank you very much. I knew it was in there, but I was too thick to pull it out. Very nice. –  Cody May 30 '11 at 9:53

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