Finding the integral of cos(ax)/sinh(bx) from 0 to infinity

[Nima]

Given Integral (Between 0 and infinity) sin ax/sinh bx = (pi/2b).tanh[(api)/(2b)]. Calculate:

Integral (Between 0 and infinity) cos ax/sinh bx dx.

Any ideas? Cheers.

 

[TenaliRaman]

Do u know leibnitz rule of differentiation under integral sign?

-- AI

 

[saltydog]

Jesus. That one neither and I spent some time on it too. Thanks.

Nima, can you report the answer please. Try and use LaTeX (look at the on-line reference for it in the group). Here, I'll start it for you, just select "quote" and add on:

$$\int_0^\infty \frac{\cos(ax)}{\sinh(bx)}=$$

 

[dextercioby]

If you differentiate wrt "a" under the integral sign,i'd like to know how one gets read o the nasty "x" that will appear in the numerator...

Daniel.

 

[saltydog]

made mistake with formula

Jesus. That one neither and I spent some time on it too. Thanks.

Nima, can you report the answer please. Try and use LaTeX (look at the on-line reference for it in the group). Here, I'll start it for you, just select "quote" and add on:

$$\int_0^\infty \frac{\cos(ax)}{\sinh(bx)}=$$

Nima, where you at? I aint' letting this one go. Just differentiate it throughout using Leibniz's rule (and no I didnt' know to do this until they told me). Granted, you have to take the derivative of infinity but just think of the upper limit as a constant 'c' and at the limit as c goes to infinity, it's derivative is still zero. Else, I'll write it up tomorrow.

This is Leibnitz's rule. Just fill in the blanks.


$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)}G(x,t)dt=G(x,\beta(x))\frac{d\beta}{dx}-G(x,\alpha(x))\frac{d\alpha}{dx}+\int_{\alpha(x)}^{\beta(x)}\frac{\partial G}{\partial x}dt$$


Salty

Edit: I made a mistake with the formula and corrected it. Now I understand what Daniel meant about the 'x' in the numerator. Will need to regroup . . .

Sorry if I caused problems for anyone.

 

[TenaliRaman]

salty,
F(a) = some integral with respect to 'x' containing parameter 'a'
F'(a) = some integral with respect to 'x' containing parameter 'a'

Can u see what i am pointing at?
My mention of Leibnitz's rule of differentiation was just to indicate that u can differentiate under integral sign, as such u don't explicitly need that formula for this problem :)

-- AI

 

[saltydog]

salty,
F(a) = some integral with respect to 'x' containing parameter 'a'
F'(a) = some integral with respect to 'x' containing parameter 'a'

Can u see what i am pointing at?
My mention of Leibnitz's rule of differentiation was just to indicate that u can differentiate under integral sign, as such u don't explicitly need that formula for this problem :)

-- AI

Alright, I'm stuck. I used Leibnitz rule incorrectly above. Next time I'll double-check things before I say anything Nima. Sorry. Gonna work on it some more and check it with real data before I post anything.

 

[saltydog]

salty,
F(a) = some integral with respect to 'x' containing parameter 'a'
F'(a) = some integral with respect to 'x' containing parameter 'a'

Can u see what i am pointing at?
My mention of Leibnitz's rule of differentiation was just to indicate that u can differentiate under integral sign, as such u don't explicitly need that formula for this problem :)

-- AI

Seems to me, after looking at it some more, that the cos integral diverges but I don't want to cause more confussion for Nima or anyone else. Would still like to find the antiderivative to prove explicitly that it does. Am I correct in this analysis?

I mean, it would have been a more interesting question if we needed to find out what $\frac{\cos(ax)}{\cosh(bx)}$ was. That's just me though.