Integrate e^(-x^2)*I_0(1-x), x from -1 to 1?

Question

Can this integral be done in closed form? If yes, then how?

Integral of e^(-x^2)*I_0(1-x), x from -1 to 1

where e is the base of the natural log and I_0 is the Bessel I function with index 0.

Thanks in advance!

Even if you can't get a result, please show your scribblings. Thanks!

Yes, I'm looking for an exact result, not a numerical one. Numerically the result is approximately 2.02848.

Answer 1

I assume you're looking for an analytical, not a numerical solution.

I didn't even know it was possible to analytically integrate the Bessel Fn I_0(x) itself.

The I_0(1-x) might be a hint to do the variable transform y=1-x but I don't see how. Setting it up as Integration by Parts doesn't seem to do anything useful either.

The e^(-x²) factor seems to be the strongest hint, not sure how it helps, but :

a) Try this identity:
J_n'(x) = J_(n-1) (x) - (n/x) J_n(x)
Then:
J_1'(x) = J_0 (x)
J_1(x) = ∫ J_0 (x) dx
looks promising...

b) To use the e^(-x²) factor, try looking at the Generating Function of the Bessel Function of the first kind:
e^(x(t-1/t)/2) = Σ_n=-∞_n=+∞ [Jn(x)*tⁿ]

c) or this definition:
I_n(x) = 1/π ∫_0_π cos (nθ - x sin θ)dθ

I_0(x) simplifies to n=0 case:
I_0(x) = 1/π ∫_0_π cos (- x sin θ)dθ
= 1/π ∫_0_π cos (x sin θ)dθ ; since cos is an even fn

Then I_0(1-x) = 1/π ∫_0_π cos ((1-x) sin θ)dθ


The Generating Function or the above should work, but if they fail, the most promising other things seem:
d) one of the many other Recurrence Relations listed below
e) numerical approximation