Approximating $\int_{-\infty}^\infty \frac{e^{-y^2/2}}{((y+y_0)^2+x_0^2)^r}\,dy$

Question

I have to estimate the integral $$\int_{-\infty}^\infty \frac{e^{-y^2/2}}{((y+y_0)^2+x_0^2)^r} \,dy,$$ for $r\in \mathbb{R}^+$. I am a little amazed that Sage and Wolfram Alpha have nothing to say about it, and that Gradshteyn-Ryzhik doesn't seem to have anything on it either; it feels like a rather natural integral, the denominator being a distance function.

Of course I realize that, if I just expand $1/((y+y_0)^2+x_0^2)^r$ into a Taylor series around $y=-y_0$ and integrate term-by-term, I am not going to get something convergent; what I get is an asymptotic formula. But what is the right order of magnitude of the error term when the formula gets cut off at the $k$th term?

For instance: let $f(y)=1/((y+y_0)^2+(x_0^2))^r$ and write $$f(y) = f(-y_0) + \frac{(y+y_0)^2}{2} O^*(\max_t f''(t)).$$ Then we get an error term of size $O(1/x_0^{2 r + 3})$. If we go up to a higher-order approximation, we obtain an error term of the form $O(1/x_0^{2(r+k)+1})$ for higher $k$. But can one also give an error term that depends on $y_0$ and not just on $x_0$, and thus is better when $x_0$ is large and $y_0$ is much larger still?

Answer 1

If you want to approximate near $y=-y_0$ you should use what you have. But for farther values of $y$ i.e. $|y+y_0|\gg0$, you should see instead that

\begin{align}\frac1{[(y+y_0)^2+x_0^2]^r}&=\frac1{(y+y_0)^{2r}}\frac1{\left[1+\left(\frac{x_0}{y+y_0}\right)^2\right]^r}\\&=\sum_{n=0}^\infty\binom{-r}n\frac{x_0^{2n}}{(y+y_0)^{2r+2n}}\end{align}

which goes to $0$ as $|y|\to\infty$.

Answer 2

Result

I have derived the following quickly converging series expansion for the integral

$$f = \sum_{k=0}^\infty a(k,x_0,y_0,r)\tag{1a}$$

where

$a(k,x_0,y_0,r) \\=(\frac{\sqrt{\pi}}{k! \Gamma(r)}) 2^{\frac{1}{2}-k} e^{-\frac{y_0^2}{2}} \left(y_0^2\right)^k \left(2^{-r} \text{B}(r,k-r+\frac{1}{2}) \, _1F_1\left(r;-k+r+\frac{1}{2};\frac{x_0^2}{2}\right)+2^{-k-\frac{1}{2}} x_0^{2 k-2 r+1} \Gamma \left(-k+r-\frac{1}{2}\right) \, _1F_1\left(k+\frac{1}{2};k-r+\frac{3}{2};\frac{x_0^2}{2}\right)\right)\tag{1b}$

where $B(x,y) =\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$ is the Euler beta function and $_1 F _1$ is a hypergeometric function.

In order to avoid typing work here is the Mathematica expression

1/(k! Gamma[r]) 2^(
 1/2 - k) E^(-(y0^2/
  2)) Sqrt[\[Pi]] (y0^2)^k (2^(-(1/2) - k) x0^(1 + 2 k - 2 r)
     Gamma[-(1/2) - k + r] Hypergeometric1F1[1/2 + k, 3/2 + k - r, 
     x0^2/2] + 
   1/Gamma[1/2 + k] 2^-r Gamma[1/2 + k - r] Gamma[
     r] Hypergeometric1F1[r, 1/2 - k + r, x0^2/2])

Numerical example:

For ${r = 1, x_0 = 1, y_0 = 1}$ the first 5 terms are
$\{0.99686, 0.261743, 0.0415358, 0.00495025, 0.000477208\}$

and their sum is $1.30557$. The numerical integration gives $fn \simeq 1.30561$.

Derivation

Here's a brief sketch of the derivation.

We start from the formula

$$z^{-r} = \frac{1}{\Gamma (r)} \int_0^{\infty } t^{r-1} \exp (-t z) \, dt\tag{2}$$

Letting $z = x_0^2+(y+y_0)^2$ and interchanging the order of integration the $y$-integral can be done exactly to give

$$\int_{-\infty }^{\infty } \exp \left(-t \left(x_0^2+(y+ y_0)^2\right)-\frac{y^2}{2}\right) \, dy\\ = \frac{\sqrt{2 \pi } e^{-t \left(\frac{y_0^2}{2 t+1}+x_0^2\right)}}{\sqrt{2 t+1}}$$

Now the $t$-integral

$$\frac{1}{\Gamma (r)} \left(\sqrt{2 \pi } e^{-\frac{y_0^2}{2}}\right) \int_0^{\infty } t^{r-1} e^{-t x_0^2}\frac{ exp({\frac{y_0^2/2}{2 t+1}})}{\sqrt{2 t+1}} \, dt$$

can't be done exactly but we can expand the exponential function containing $y_0^2 \frac{1}{2t+1}$ in a power series, and then the coefficients can be integrated to give $(1)$.

Special cases

The case $r=0$

gives $f(0,x_0,y_0) = \sqrt{2 \pi}$

The case $y_0 = 0$

$f(r,x_0,y_0=0) = 2^{\frac{1}{2}-r} \Gamma \left(\frac{1}{2}-r\right) \, _1F_1\left(r;r+\frac{1}{2};\frac{x_0^2}{2}\right)+\left(x_0^2\right)^{\frac{1}{2}-r} B\left(\frac{1}{2},r-\frac{1}{2}\right) \, _1F_1\left(\frac{1}{2};\frac{3}{2}-r;\frac{x_0^2}{2}\right)$

Asmptotic behaviour

$f(r,x_0 \simeq 0,y_0=0) \simeq \left(x_0^2\right)^{\frac{1}{2}-r} B\left(\frac{1}{2},r-\frac{1}{2}\right)+2^{\frac{1}{2}-r} \Gamma \left(\frac{1}{2}-r\right) +O\left(x_0^2\right)$

$f(r,x_0 \simeq \infty,y_0=0) \simeq \sqrt{2\pi} x_0^{-2r}$

Answer 3

Mainly for purposes of comparison, let me give a bound using what amounts to a cheap version of Laplace's method. (Cross-posted on MathOverflow.)

Choose $\rho\in (0,1)$. Let $g(y) = 1/(x_0^2+y^2)^{\sigma/2}$. Then $$g''(y) = \frac{-\sigma}{(x^2+y^2)^{\frac{\sigma}{2} + 1}} + \frac{\sigma \left(\frac{\sigma}{2}+1\right) \cdot 2 y^2}{(x^2+y^2)^{\frac{\sigma}{2} + 2}}$$ and so $|g''(y)|\leq \sigma (\sigma+1)/(x^2+y^2)^{\sigma/2+1}$. Let $I$ be the interval $\lbrack (1-\rho) y_0, (1+\rho) y_0\rbrack$. Then, for $y\in I$, $|g''(y)|\leq \sigma (\sigma+1)/((1-\rho) l_0)^{\sigma+2}$, where $l_0 = \sqrt{x_0^2+y_0^2}$, and so $$g(y) = g(y_0) + g'(y_0) (y- y_0) + O^*\left(c_0 (y-y_0)^2\right),$$ where $c_0 = \sigma (\sigma+1)/2 ((1-\rho) l_0)^{\sigma+2}$. Thus, by cancellation, $$\int_I g(y) e^{-(y-y_0)^2/2} dy = \int_I g(y_0) e^{-(y-y_0)^2/2} dy + O^*\left(\int_I c_0 (y-y_0)^2 e^{-(y-y_0)^2/2} dy\right).$$ Since $g'(y)<0$ for $y\geq 0$, we also know that $g(y)<g(y_0) + c_0 (y-y_0)^2$ for $y>(1+\rho) y_0$. We conclude that $$\begin{aligned}\int_{(1-\rho) y_0}^{\infty} g(y) e^{-(y-y_0)^2/2} dy &\leq g(y_0) \int_{-\infty}^\infty e^{-y^2/2} dy + c_0 \int_{-\infty}^\infty y^2 e^{-y^2/2} dy\\ &= g(y_0) \sqrt{2\pi} + c_0 \sqrt{2\pi} = \left(1 + \frac{\sigma (\sigma+1)}{2 (1-\rho)^{\sigma+2} l_0^2}\right) \frac{\sqrt{2\pi}}{l_0^\sigma} .\end{aligned}$$

It remains to consider $y\leq (1-\rho) y_0$. Since $g(y)\leq g(0) = 1/x_0^\sigma$, $$\begin{aligned}\int_{-\infty}^{(1-\rho) y_0} g(y) e^{-(y-y_0)^2/2} dy &= \frac{1}{x_0^\sigma} \int_{-\infty}^{-\rho y_0} e^{-y^2/2} dy \\ &\leq \frac{1}{x_0^\sigma (\rho y_0)} \int_{-\infty}^{-\rho y_0} y e^{-y^2/2} dy = \frac{e^{-\rho^2 y_0^2/2}}{\rho x_0^\sigma y_0}. \end{aligned}$$

Thus we obtain $$\int_{-\infty}^\infty \frac{e^{-(y-y_0)^2/2}}{(x_0^2+y^2)^{\sigma/2}} dy \leq \left(1 + \frac{\sigma (\sigma+1)}{2 (1-\rho)^{\sigma+2} l_0^2}\right) \frac{\sqrt{2\pi}}{l_0^\sigma} + \frac{e^{-\rho^2 y_0^2/2}}{\rho x_0^\sigma y_0}$$ for any $0<\rho<1$.

Hardly very powerful or elegant, but I wonder: (a) is the above qualitatively optimal? That is, are the lesser-order terms of the right order? (b) can one give an even quicker proof of the same or a stronger bound?