Series expansion of integral

Question

Consider the function $I(y)=\int_0^\infty e^{-\sqrt{x^2+y^2}} \mathrm{d} x$. I'd like to see the leading order term of $I(y)$ about $y=0$, so I expand the integrand: $$ e^{-\sqrt{x^2+y^2}}=e^{-x}-e^{-x}\frac{1}{2x}y^2+e^{-x}\frac{1+x}{8x^3}y^4+\dots $$ However, only the integral of the $0^\mathrm{th}$ order term converges (to $1$ of course). How would one proceed here to find the second order term (or higher)?

Answer 1

As simple hyperbolic substitution of the form $x=y\sinh t$ yields $I(y)=|y|~K_1\Big(|y|\Big),$ see Bessel function for more information. Its series expansions around the origin can be found here.

Answer 2

Let $I(y)$ be the integral of interest given by

$$\begin{align} I(y)&=\int_0^\infty e^{-\sqrt{x^2+y^2}}\,dx\\\\ &=\int_0^{|y|} e^{-\sqrt{x^2+y^2}}\,dx+\int_{|y|}^\infty e^{-\sqrt{x^2+y^2}}\,dx \tag 1 \end{align}$$

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For the first integral on the right-hand side of $(1)$, we note that $\sqrt{x^2+y^2}\le \sqrt{2}|y|$. Therefore, inasmuch as we are developing an asymptotic series for "small" $|y|$, we can write

$$\begin{align} \int_0^{|y|} e^{-\sqrt{x^2+y^2}}\,dx&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^{|y|}\left(x^2+y^2\right)^{n/2}\,dx\\\\ &=\sum_{n=0}^\infty \frac{(-1)^n|y|^{n+1}}{n!}\int_0^{\pi/4}\sec^{n+2}\theta \,d\theta\\\\ &=|y|-\left(\frac{\sqrt 2 +\log\left(1+\sqrt 2\right)}{2}\right)|y|^2+\frac23 |y|^3+O\left(|y|^4\right) \tag 2 \end{align}$$

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To facilitate evaluation of the second integral on the right-hand side of $(1)$, we expand the exponential as

$$\begin{align} e^{-\sqrt{x^2+y^2}}&=e^{-x}-\frac12 y^2\frac{e^{-x}}{x}+\frac18 y^4\frac{(x+1)e^{-x}}{x^3}+O\left(y^6\right)\\\\ &=e^{-x}-\frac12 y^2\frac{e^{-x}}{x}+\frac18 y^4\frac{e^{-x}}{x^2}+\frac18 y^4\frac{e^{-x}}{x^3}+O\left(y^6\right) \tag 3\\\\ \end{align}$$

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Next, we write the integral of the first term on the right-hand side of $(3)$ as

$$\begin{align} \int_{|y|}^\infty e^{-x}\,dx&=e^{-|y|}\\\\ &=\sum_{n=0}\frac{(-1)^n|y|^n}{n!}\\\\ &=1-|y|+\frac12 |y|^2-\frac16 |y|^3+O\left(|y|^4\right) \tag 4 \end{align}$$

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We write the integral of the second term is

$$\begin{align} -\frac12 |y|^2\,\int_{|y|}^\infty \frac{e^{-x}}x\,dx&=-\frac12 |y|^2\,\left(\int_{|y|}^1 \frac{e^{-x}}x\,dx+\int_1^\infty \frac{e^{-x}}x\,dx\right)\\\\ &=\frac12 \log (|y|)\,|y|^2-\frac12 \left(\int_1^\infty \frac{e^{-x}}x\,dx-\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\,n!}\right)\,|y|^2-\frac12 \sum_{n=1}^\infty \frac{(-1)^{n-1}|y|^{n+2}}{n\,n!}\\\\ &=\frac12 \log (|y|)\,|y|^2+\frac12 \gamma\,|y|^2-\frac12 \sum_{n=1}^\infty \frac{(-1)^{n-1}|y|^{n+2}}{n\,n!}\\\\ &=\frac12 \log (|y|)\,|y|^2+\frac12 \gamma\,|y|^2-\frac12 |y|^3+O\left(|y|^4\right) \tag 5 \end{align}$$

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Using integration by parts, we find the third integral on the right-hand side of $(3)$ as

$$\begin{align} \frac18 |y|^4\,\int_{|y|}^\infty \frac{e^{-x}}{x^2}\,dx&=\frac18 |y|^4\left(\frac{e^{-|y|}}{|y|}\right)-\frac18 |y|^4\,\int_{|y|}^\infty \frac{e^{-x}}{x}\,dx\\\\ &=\frac18 e^{-|y|}|y|^3+\log (|y|)|y|^4+\frac18 \gamma |y|^4-\frac18 \sum_{n=1}^\infty \frac{(-1)^{n-1}|y|^{n+4}}{n\,n!}\\\\ &=\frac18 |y|^3+O\left(\log(|y|)\,|y|^4\right) \tag 6 \end{align}$$

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Using integration by parts twice, we find the fourth integral on the right-hand side of $(3)$ as

$$\frac18\,|y|^4\int_{|y|}^\infty \frac{e^{-x}}{x^3}\,dx=-\frac1{16}|y|^2-\frac18|y|^3+O\left(\log(|y|)\,|y|^4\right)\tag 7$$

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Putting together results from $(2)$ and $(4)-(7)$, we obtain so far

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty e^{-(x^2+y^2)}\,dx=1+\left(\frac{\log(|y|)+\gamma -\log(1+\sqrt 2)-(1-\sqrt 2)}{2}\right)\,y^2+O(y^4)}\tag 8$$

IMPORTANT NOTE:

In $(3)$, we have omitted terms beyond $\frac18 y^4\frac{e^{-x}}{x^3}$. These terms will have a component that is of order $|y|^2$ and as such, the expansion in $(8)$ as written, is not correct.

However, the expansion is correct up to order $\log(|y|)\,|y|^2$. On way to complete the expansion up to order $|y|^2$ is to use the Faa di Bruno Formula to write the full expansion of $(3)$.

Answer 3

If $x^2+y^2 =r^2 $, $2x dx =2r dr $ or $dx =\frac{r dr}{x} =\frac{r dr}{\sqrt{r^2-y^2}} $

$\begin{align} I(y) &=\int_0^\infty e^{-\sqrt{x^2+y^2}}\,dx\\ &=\int_y^\infty e^{-r}\frac{r dr}{\sqrt{r^2-y^2}}\\ &=\int_1^\infty e^{-ys}\frac{ys y ds}{\sqrt{y^2s^2-y^2}} \qquad r = ys, dr = yds\\ &=y\int_1^\infty e^{-ys}\frac{s ds}{\sqrt{s^2-1}}\\ &=y\int_1^\infty e^{-ys}\frac{ ds}{\sqrt{1-1/s^2}}\\ &=y\int_1^\infty e^{-ys}(1-1/s^2)^{-1/2}ds\\ &=y\int_1^\infty e^{-ys}\sum_{n=0}^{\infty}(-1)^n\binom{-1/2}{n}\frac{ds}{s^{2n}}\\ &=y\sum_{n=0}^{\infty}(-1)^n\binom{-1/2}{n}\int_1^\infty e^{-ys}s^{-2n}ds\\ &=y\sum_{n=0}^{\infty}(-1)^n\binom{-1/2}{n}y^{2n-1}\int_1^\infty e^{-ys}(ys)^{-2n}yds\\ &=\sum_{n=0}^{\infty}(-1)^n\binom{-1/2}{n}y^{2n}\int_y^\infty e^{-t}t^{-2n}dt\\ &=\sum_{n=0}^{\infty}(-1)^n\binom{-1/2}{n}y^{2n}\Gamma(-2n+1,y)\\ \end{align} $

From http://dlmf.nist.gov/8.4 we have $\Gamma(-n,z) =\frac{(-1)^{n}}{n!}(E_{1}(z)-e^{-z} \sum_{k=0}^{n-1}\frac{(-1)^{k}k!}{z^{k+1}}) $ where $E_{1}(z) =\int_{z}^{\infty}\frac{e^{-t}}{t}dt $ so

$\begin{array}\\ y^{2n}\Gamma(-2n+1,y) &=y^{2n}\frac{(-1)^{2n-1}}{(2n-1)!}(E_{1}(z)-e^{-z} \sum_{k=0}^{2n-2}\frac{(-1)^{k}k!}{y^{k+1}})\\ &=-\frac{1}{(2n-1)!}(y^{2n}E_{1}(y)-e^{-y} \sum_{k=0}^{2n-2}(-1)^{k}k!y^{2n-k-1})\\ &=-\frac{1}{(2n-1)!}(y^{2n}E_{1}(y)-e^{-y} \sum_{k=0}^{2n-2}(-1)^{2n-2-k}(2n-2-k)!y^{2n-(2n-2-k)-1})\\ &=-\frac{1}{(2n-1)!}(y^{2n}E_{1}(y)-e^{-y} \sum_{k=0}^{2n-2}(-1)^{k}(2n-2-k)!y^{k+1})\\ \end{array} $

Since $E_{1}(z) \sim\frac{e^{-z}}{z}\left(1-\frac{1!} {z}+\frac{2!}{z^{2}}-\frac{3!}{z^{3}}+\cdots\right) $, we have $y^{2n}E_{1}(y) \sim y^{2n-1}e^{-y}\left(1-\frac{1!} {y}+\frac{2!}{y^{2}}-\frac{3!}{y^{3}}+\cdots\right) $.

Also, since $(-1)^n\binom{-1/2}{n} =\binom{2n}{n}\frac{1}{2^{2n+1}} $, we finally get

$\begin{array}\\ I(y) &=\sum_{n=0}^{\infty}(-1)^n\binom{-1/2}{n}y^{2n}\Gamma(-2n+1,y)\\ &=-\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{2^{2n+1}}\left(\frac{1}{(2n-1)!}\left(y^{2n}E_{1}(y)-e^{-y} \sum_{k=0}^{2n-2}(-1)^{k}(2n-2-k)!y^{k+1}\right)\right)\\ &\sim-\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{2^{2n+1}}\left(\frac{1}{(2n-1)!}\left(y^{2n-1}e^{-y}-e^{-y} \sum_{k=0}^{2n-2}(-1)^{k}(2n-2-k)!y^{k+1}\right)\right)\\ &=-e^{-y}\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{2^{2n+1}(2n-1)!}\left(y^{2n-1}- \sum_{k=0}^{2n-2}(-1)^{k}(2n-2-k)!y^{k+1}\right)\\ \end{array} $

with the last approximation using $y^{2n}E_{1}(y) \sim y^{2n-1}e^{-y} $.

As usual, with this complexity, $P(error) > 1-\frac1{e} $, but I think that something like this should hold.