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Topic: integration question: 1/(Sqrt[1 - x^2])*Exp[-(1 - x)^2]
Replies: 4   Last Post: Aug 5, 2013 5:51 PM

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Nasser Abbasi

Posts: 5,686
Registered: 2/7/05
Re: integration question: 1/(Sqrt[1 - x^2])*Exp[-(1 - x)^2]
Posted: Aug 5, 2013 4:43 PM
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On 8/5/2013 3:26 PM, Nasser M. Abbasi wrote:
> On 8/5/2013 3:11 PM, Axel Vogt wrote:
>> On 05.08.2013 21:58, Nasser M. Abbasi wrote:
>>> 1/(Sqrt[1 - x^2])*Exp[-(1 - x)^2]
>>
>> Plot from 0 ... 1 and its imaginary part from 1 ... 4
>> to understand the the problem
>>

>
> Oh, a pole at x=1 ofcourse.
>
> This explains why numerical integration worked
> (but lots of screaming from Mathematica along the way)
>
> NIntegrate[1/(Sqrt[1 - x^2])*Exp[-(1 - x)^2], {x, 0, Infinity}]
>
> 1.29331348557849 - 1.19124235143936 I
>
> same with Maple's
>
> int(1/(sqrt(1 - x^2 ))*exp((- (1 - x)^2)),x=-0..infinity,numeric=true);
> 1.29721962103293 - 1.19514845220774 I
>
> Ok. Sorry, I just overlooked this.
>


Just a quick follow up. Since

Limit[1/(Sqrt[1 - x^2])*Exp[-(a - x)^2], x -> a]

gives

1/Sqrt[1 - a^2]

and so when a=1 there is a pole. Then why assuming a /= 1 still
does not produce any result?

Assuming[a!=1,Integrate[(1/Sqrt[1-x^2])*Exp[-(a - x)^2], {x, 0, Infinity}]]

Tried many other assumptions, and non working.

Now the pole is elminated. But the new problem, it becomes complex integration
now.

Any way to proceed now? Does one need to do complex contour integration?

thank you,
--Nasser









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