Integral (related to Laplace transform)

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SUMMARY

The integral discussed, \int\limits_0^{ + \infty } {\frac{{dt}}{{{e^{st}} \cdot (1 + {e^t})}}}, is identified as a non-elementary integral. By substituting u = e^t, the integral transforms into J = \int_1^{\infty} \frac{u^{-s-1}}{1+u} \, du, which is evaluated using Maple 11 as J = -\text{LerchPhi}(-1,1,-s) - \pi \csc(\pi s). The LerchPhi function is defined for complex variables and is extended through analytic continuation. The discussion highlights the challenges of integrating this expression, particularly in the context of differential equations.

PREREQUISITES
  • Understanding of Laplace transforms and their properties
  • Familiarity with non-elementary integrals
  • Knowledge of the LerchPhi function and its applications
  • Proficiency in using Maple 11 for symbolic computation
NEXT STEPS
  • Study the properties and applications of the LerchPhi function
  • Learn about non-elementary integrals and their evaluation techniques
  • Explore the use of Maple 11 for advanced integral calculations
  • Investigate convolution methods in the context of differential equations
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Mathematicians, engineers, and students working on advanced calculus, particularly those dealing with Laplace transforms and differential equations.

Pzi
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Hi.

Homework Statement


[tex]\int\limits_0^{ + \infty } {\frac{{dt}}{{{e^{st}} \cdot (1 + {e^t})}}}[/tex]

Homework Equations


The same as...
[tex]Laplace\left[ {\frac{1}{{1 + {e^t}}}} \right][/tex]

The Attempt at a Solution


Found no elegant properties related to Laplace transform here.
So figured my best shot would be to integrate directly...

Any suggestions?
 
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Pzi said:
Hi.

Homework Statement


[tex]\int\limits_0^{ + \infty } {\frac{{dt}}{{{e^{st}} \cdot (1 + {e^t})}}}[/tex]

Homework Equations


The same as...
[tex]Laplace\left[ {\frac{1}{{1 + {e^t}}}} \right][/tex]

The Attempt at a Solution


Found no elegant properties related to Laplace transform here.
So figured my best shot would be to integrate directly...

Any suggestions?

I think it is a non-elementary integral. By changing variables to u = exp(t), and recognizing that for s > 0 we have exp(-s*t) = exp(-s*ln(u)) = u^(-s), the integral becomes
[tex]J = \int_1^{\infty} \frac{ u^{-s-1}}{1+u} \, du,[/tex] which Maple 11 evaluates as
[tex]J = -\text{LerchPhi}(-1,1,-s) - \pi \csc(\pi s),[/tex]
where LerchPhi is the function defined as
[tex]\text{LerchPhi}(z,a,v) = \sum_{n=0}^{\infty} \frac{z^n}{(v+n)^a}[/tex] if [itex]|z| < 1[/itex] or [itex]|z|=1[/itex] and [itex]\text{Re}(a) > 1.[/itex] It is extended to the whole complex z-plane by analytic continuation.

RGV
 
Ray Vickson said:
I think it is a non-elementary integral. By changing variables to u = exp(t), and recognizing that for s > 0 we have exp(-s*t) = exp(-s*ln(u)) = u^(-s), the integral becomes
[tex]J = \int_1^{\infty} \frac{ u^{-s-1}}{1+u} \, du,[/tex] which Maple 11 evaluates as
[tex]J = -\text{LerchPhi}(-1,1,-s) - \pi \csc(\pi s),[/tex]
where LerchPhi is the function defined as
[tex]\text{LerchPhi}(z,a,v) = \sum_{n=0}^{\infty} \frac{z^n}{(v+n)^a}[/tex] if [itex]|z| < 1[/itex] or [itex]|z|=1[/itex] and [itex]\text{Re}(a) > 1.[/itex] It is extended to the whole complex z-plane by analytic continuation.
RGV
Yea, looks like it's indeed a non-elementary case here...
As this is a part of some differential equations I will work around it (using convolution), but it will be ugly.

Thanks for clarification!
 

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