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 November 8th, 2012, 02:10 AM #1 Newbie   Joined: May 2011 Posts: 22 Thanks: 0 Does this integration have a close-form result? Hi guys, I have encountered a problem, the integral of following expression from 0 to $\infty$: $\int_0^\infty \frac{1 - \exp(-C-D\cdot B^x)}{1+A\cdot B^x} dx$ where $A>0,C>0, D>0$ are all positive constants, and $B>1$. Not sure there is close-form result for this integration. Maybe using Gamma function is also OK. Thanks for any solutions or hints!!
November 8th, 2012, 05:30 AM   #2
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Re: Does this integration have a close-form result?

Quote:
 Originally Posted by jackson_wang Not sure there is close-form result for this integration.
I think there aren't any. Try using CA.

 November 8th, 2012, 03:49 PM #3 Newbie   Joined: May 2011 Posts: 22 Thanks: 0 Re: Does this integration have a close-form result? Hi, What is CA? Thanks. Jackson
 November 8th, 2012, 10:00 PM #4 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Does this integration have a close-form result? Complex Analysis. Use residue theorem.
 November 8th, 2012, 11:20 PM #5 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Does this integration have a close-form result? I believe that this integral can be solved by hand using incomplete Gamma function.
November 9th, 2012, 01:58 AM   #6
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Re: Does this integration have a close-form result?

Quote:
 Originally Posted by zaidalyafey I believe that this integral can be solved by hand using incomplete Gamma function.

Hi zaidalyafey,

Could you show me how to use Gamma function to represent this? Or any detailed hint.
I appreciate it sincerely.

Jackson

 November 9th, 2012, 02:35 AM #7 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Does this integration have a close-form result? $\int_0^\infty \frac{1 - \exp(-C-D\cdot B^x)}{1+A\cdot B^x} dx$ The first part we solved it viewtopic.php?f=15&t=35491 $\text{Now for : }\int_0^\infty \frac{- \exp(-C-D\cdot B^x)}{1+A\cdot B^x} dx$ $=\frac{-1}{e^{C}}\int_0^\infty \frac{ dx}{e^{D\cdot B^x}\cdot (1+A\cdot B^x)}$ $\text{Now use substitution : }t=AB^x\Rightarrow \,\,\log t \,=\, \log A\,+\,x\log B\Rightarrow \frac{dt}{t}\,=\, \log B\, dx$ $=\frac{-1}{e^{C}\cdot \log B}\,\int_D^\infty\, \frac{ dt}{\,e^{$$\frac{D\cdot t}{A}$$}\cdot t(1+t)}\text{ By partial fraction : }$ $=\frac{-1}{e^{C}\cdot \log B}\,$\int_D^\infty\, \frac{ dt}{\,e^{$$\frac{D\cdot t}{A}$$}\cdot t}\,-\,\frac{ dt}{\,e^{$$\frac{D\cdot t}{A}$$}\cdot (1+t)}$$ I hope you can continue from here .
November 11th, 2012, 07:12 PM   #8
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Re: Does this integration have a close-form result?

Quote:
 Originally Posted by zaidalyafey $\int_0^\infty \frac{1 - \exp(-C-D\cdot B^x)}{1+A\cdot B^x} dx$ The first part we solved it viewtopic.php?f=15&t=35491 $\text{Now for : }\int_0^\infty \frac{- \exp(-C-D\cdot B^x)}{1+A\cdot B^x} dx$ $=\frac{-1}{e^{C}}\int_0^\infty \frac{ dx}{e^{D\cdot B^x}\cdot (1+A\cdot B^x)}$ $\text{Now use substitution : }t=AB^x\Rightarrow \,\,\log t \,=\, \log A\,+\,x\log B\Rightarrow \frac{dt}{t}\,=\, \log B\, dx$ $=\frac{-1}{e^{C}\cdot \log B}\,\int_D^\infty\, \frac{ dt}{\,e^{$$\frac{D\cdot t}{A}$$}\cdot t(1+t)}\text{ By partial fraction : }$ $=\frac{-1}{e^{C}\cdot \log B}\,$\int_D^\infty\, \frac{ dt}{\,e^{$$\frac{D\cdot t}{A}$$}\cdot t}\,-\,\frac{ dt}{\,e^{$$\frac{D\cdot t}{A}$$}\cdot (1+t)}$$ I hope you can continue from here .

Hi @zaidalyafey

Thanks a lot for this.
I can see the former part of your last equation i.e., $\int_A^\infty \frac{ dt}{\,e^{$$Bt$$}\cdot t} dt$
can be expressed by incomplete Gamma function.

However, just wondering the latter part, i.e., $\int_A^\infty \frac{ dt}{\,e^{$$Bt$$}\cdot (1+t)} dt$
can be represented similarly?

Thanks a lot.

 November 12th, 2012, 01:30 AM #9 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Does this integration have a close-form result? $\text{Assume we have : }\int^{\infty}_{R}\,\frac{dx}{e^x \cdot (1+x)}$ $\text{Use a substitution : }1+x=t$ $\text{Then this becomes : }e\int^{\infty}_{R+1}\,\frac{dt}{t\, \cdot e^{t\,}}$
 November 12th, 2012, 01:42 AM #10 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Does this integration have a close-form result? $\text{You can also use the exponential integeral : }$ $\mbox{Ei}(x)=\int_{-\infty}^x\frac{e^t}t\,dt.$

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