What is the value of this integral?

[rsq_a]

There is a paper that compares numerical methods of calculating certain principal value integrals. One of them is,

\displaystyle P \int_0^1 \frac{e^{-y^2}}{y-0.25} dy

The author has written down a (numerically) exact value of 0.438392.

Does anybody know how he got that? As far as I know, there is no representation of that integral in terms of special functions. Mathematica doesn't give me anything (numerical or otherwise), and Gradshteyn and Ryzhik's Table of Integrals doesn't seem to have it.

 

[CompuChip]

It looks to me like there's quite a nasty singularity at y = 0.25 (decimal numbers :() so I don't even see how this converges.

 

[rsq_a]

It looks to me like there's quite a nasty singularity at y = 0.25 (decimal numbers :() so I don't even see how this converges.

It's a principal value integral. I wrote it in the intro. But I'll clarify the latex.

 

[CompuChip]

Whoops, I completely missed that. If I push Mathematica to its precision limits, I get approximately the same answer but I don't get the decimals right. I don't see any way of evaluating this analytically either...

 

[rsq_a]

Whoops, I completely missed that. If I push Mathematica to it's precision limits, I get approximately the same answer but I don't get the decimals right. I don't see any way of evaluating this analytically either...

Yes, Maple can do it, too.

I doubt there is an analytical answer. What I assumed the author had was an expression in terms of special functions -- which could be 'exactly' calculated via tables -- or at least some numerical method not limited by the singularity.

Odd though. The other functions he looks at are,

\int_0^1 \frac{y^6}{y-0.25} dy
\int_0^1 \frac{\cos{y}}{y-0.25} dy
\int_0^1 \frac{exp^{-y}}{y-0.25} dy

And each of these have exact answers (in terms of special functions in the second and third cases).