8
$\begingroup$

I am trying to evaluate the integral \begin{align} \frac{1}{2\sqrt{2}\pi}\int_{0^{-}}^{t} ds \ \frac{e^{-x^2/2S^2(t,s) }}{\Sigma(s) S(t,s)} \end{align} where $S(t, s) = 2D(t-s)+\frac{\Sigma(s)}{2}$ and $\Sigma(s)= \sigma^2+2Ds$. I found a rather neat change of variable by taking $\xi=S^{-1}(t,s)$ so that \begin{align} \Sigma(s)&=\sqrt{2} \ \xi^{-1}\sqrt{\xi^2\Sigma^2(t)-1}\\ 2D(t-s)&=2 \ \xi^{-2}-\Sigma^2(t)\\ ds&=\frac{2}{D\xi^3}\ d\xi \end{align} Applying this to the integral gives the result \begin{align} \frac{1}{2 \pi D}\int_{\sqrt{2}/\Sigma (2 t)}^{\sqrt{2}/\Sigma (t)} d{\xi} \ \frac{e^{-x^2 \ \xi^2/2 }}{\xi\sqrt{\Sigma^2(t)\xi^2-1}}=\frac{1}{2 \pi D}{\int_{\xi_L}^{\xi_H} d{\xi} \frac{e^{-x^2 \ \xi^2/2 }}{\xi\sqrt{\Sigma^2(t)\xi^2-1}}}. \end{align} and at last applying the change of variable $y=\Sigma(t)\xi$ gives \begin{align} \frac{1}{2 \pi D}{\int_{\xi_L}^{\xi_H} d{\xi} \frac{e^{-x^2 \ \xi^2/2 }}{\xi\sqrt{\Sigma^2(t)\xi^2-1}}} = \frac{1}{2 \pi D}{\int_{\sqrt{2}\Sigma(t)/\Sigma(2t)}^{\sqrt{2}} d{y} \frac{e^{-(x^2/2\Sigma^2(t)) y^2 }}{y\sqrt{y^2-1}}} =\frac{1}{2 \pi D}\color{blue}{\int_{a}^{b} d{y} \frac{e^{-c y^2 }}{y\sqrt{y^2-1}}} \end{align} Above calculation indeed check out numerically. enter image description here It seems the integral in blue could have an anti-derivative. However, no luck just yet! Someone who knows a way forward? Many thanks in advance!

$\endgroup$
6
  • $\begingroup$ Evaluating the integral , there is nothing to solve $\endgroup$
    – jimjim
    Aug 13 '21 at 9:07
  • $\begingroup$ There is no closed form even in the case $a=0$ and $b=\infty$ as the integral becomes the Laplace transform of $1/(t\sqrt{\sqrt t-1})$. $\endgroup$
    – TheSimpliFire
    Aug 13 '21 at 10:24
  • $\begingroup$ @TheSimpliFire thanks for the answer! Do you have a prove or source to support the claim? Or do you know of a method to prove an integral has no closed form? Many thanks in advance! $\endgroup$
    – drandran12
    Aug 13 '21 at 11:22
  • 2
    $\begingroup$ @drandran12 By no closed form I mean it can't be represented using elementary functions. Another way to look at it is to rewrite the integral problem as the solution of the DE $f'(x)=e^{-cx^2}/(x\sqrt{x-1})$. Writing $f(x)=e^{-cx^2}\sqrt{x-1}g(x)$ yields the ODE $$(x^2-x)g'(x)+(-2cx^3+2cx^2+x/2)g(x)=1$$ where the high-order polynomial terms strongly indicate a lack of a closed form (even more so when $c$ is arbitrary). $\endgroup$
    – TheSimpliFire
    Aug 13 '21 at 12:24
  • $\begingroup$ @TheSimpliFire I know what you meant by "no closed form" :) However, why does the high-order polynomial terms strongly indicate a lack of a closed form? Sorry for the maybe stupid question :/ $\endgroup$
    – drandran12
    Aug 13 '21 at 13:35
4
+50
$\begingroup$

As others have pointed out, no closed-form solution in terms of elementary functions can be found. However, if you can live with the error function and the Owen T function then a closed-form solution can be found in terms of these special functions.

I will assume the error function $\operatorname{erf} (x)$ is well known to you. An integral representation for the Owen T function we intent to use can be found here. It is $$\operatorname{T}(h,a) = \frac{1}{2\sqrt{2\pi}} \int_h^\infty e^{-x^2/2} \operatorname{erf} \left (\frac{ax}{\sqrt{2}} \right )\, dx.$$ Enforcing a substitution of $x \mapsto x \sqrt{2}$ gives the following form that will be found most convenient for the integral considered here: $$\int_z^\infty e^{-x^2} \operatorname{erf}(ax) \, dx = 2 \sqrt{\pi} \operatorname{T}(z \sqrt{2}, a). \tag1$$

Now let us look at the integral in question. Let $$I(t) = \int_a^b \frac{e^{-tx^2}}{x\sqrt{x^2 - 1}} \, dx, \quad t > 0.$$ I will assume $1 < a < b = \sqrt{2}$. Note that $I(\infty) = 0$ and we are required to find $I(c)$ where $c > 0$. Using Feynman's trick of differentiating under the integral sign with respect to $t$ gives $$I'(t) = - \int_a^b \frac{x e^{-tx^2}}{\sqrt{x^2 - 1}} \, dx.$$ Enforcing a substitution of $u^2 = x^2 - 1$ yields \begin{align*} I'(t) &= e^{-t} \int_{\alpha_1}^{\alpha_2} e^{-tu^2} \, du = -\frac{\sqrt{\pi} e^{-t}}{2\sqrt{t}} \operatorname{erf}(u\sqrt{t}) \Big{|}_{\alpha_1}^{\alpha_2}\\ &= -\frac{\sqrt{\pi} e^{-t}}{2\sqrt{t}} \left [\operatorname{erf}(\alpha_2 \sqrt{t}) - \operatorname{erf}(\alpha_1 \sqrt{t}) \right ]. \end{align*} Here $\alpha_1 = \sqrt{a^2 - 1}$ and $\alpha_2 = \sqrt{b^2 - 1}$.

Now, as $$\int_c^\infty I'(t) \, dt = I(\infty) - I(c) = -I(c),$$ since $I(\infty) = 0$, we have $$I(c) = -\frac{\sqrt{\pi}}{2} \int_c^\infty \frac{e^{-t}}{\sqrt{t}} \left [\operatorname{erf} \left (\alpha_2 \sqrt{t} \right ) - \operatorname{erf} \left (\alpha_1 \sqrt{t} \right ) \right ] \, dt.$$ On enforcing a substitution of $t \mapsto t^2$ one obtains $$I(c) = \sqrt{\pi} \int_{\sqrt{c}}^\infty e^{-t^2} \left [\operatorname{erf}(\alpha_2 t) - \operatorname{erf}(\alpha_1 t) \right ] \, dt.$$ Evaluating this integral in terms of the Owen T function, from (1) one immediately sees that $$I(c) = 2\pi \left (\operatorname{T} \left (\sqrt{2c}, \alpha_2 \right ) - \operatorname{T} \left (\sqrt{2c}, \alpha_1 \right ) \right ),$$ or $$\int_a^b \frac{e^{-cy^2}}{y\sqrt{y^2 - 1}} \, dy = 2\pi \left (\operatorname{T} \left (\sqrt{2c}, \sqrt{b^2 - 1} \right ) - \operatorname{T} \left (\sqrt{2c}, \sqrt{a^2 - 1} \right ) \right ),$$ the required closed-form expression for our integral.

Note, if $b = \sqrt{2}$, as it appears to be from the question, further simplification of one of the Owen T functions is possible. As can be seen here when the right argument of the Owen T function is unity $$\operatorname{T} \left (\sqrt{2c}, \sqrt{b^2 - 1} \right ) = \operatorname{T} \left (\sqrt{2c}, 1 \right ) = \frac{1}{8} \left [1 - \operatorname{erf}^2 \left (\sqrt{c} \right ) \right ],$$ allowing one to write $$\int_a^{\sqrt{2}} \frac{e^{-cy^2}}{y\sqrt{y^2 - 1}} \, dy = -\frac{\pi}{4} \left [8 \operatorname{T} \left (\sqrt{2c}, \sqrt{a^2 - 1} \right ) + \operatorname{erf}^2 \left (\sqrt{c} \right ) - 1 \right ].$$

$\endgroup$
1
  • $\begingroup$ Wow, thank you very much! As you can see from my previous questions, I am very familiar with the error and Owen T function. So I am very happy with the solution you found! $\endgroup$
    – drandran12
    Aug 21 '21 at 11:11
2
$\begingroup$

If $$f(x)=\frac{e^{-x^2}}{x\sqrt{x^2-1}}$$ then the integral $I=\int_a^bf(x)dx, b>a$ is not only convergent for all $a>a_0$ where $a_0$ is the root of $1/x=f(x)$, but also $f(x)\in[0,1)$, which gives a hope of some closed form of the definite integral. At least, a very close approximant. Like, the transformation $x=\sec t$ gives for the integral $$\int_\alpha^\beta e^{-\sec^2t}dt$$. A silver lining of hope.

EDIT

If we consider $E(x)=\int_0^x e^{-\sec^2t}dt$, although the Taylor expansion is a little hopeless: $$ e^{-\sec^2x} = \frac{1}{e} \left( 1 - x^2 - \frac{1}{6}x^4 + \frac {11}{90}x^6+O (x^8)\right) $$ The denominators are the sequence $$a_n = \frac{(2n)!}{2^n} $$, and I'm not sure of the numerators. Yet, via some graphing it can be seen that the following is approximately correct. $$ e^{-\sec^2x} \approx e^{-1-x^2}(1-\alpha x^4), \text{where } \alpha \in \left[ \frac{2}{3},\frac{5}{6} \right] $$ Therefore, a crude approximation, $$E(x)=\frac {\sqrt\pi}{8e}(4-3\alpha) \text{erf} (x) + e^{-1-x^2} \left( \frac{x^3}{2} +\frac{3x}{4} \right) $$

$\endgroup$
1
  • $\begingroup$ The integral of $e^{-\sec^2t}$ isn't elementary, as far as I'm aware. $\endgroup$ Aug 19 '21 at 1:58
1
$\begingroup$

The integral in question doesn't have an elementary anti-derivative for all $c$. For example, consider the case where $c=1$:

$$I = \int_1^{\sec{x}} \dfrac{e^{-y^2}}{y\sqrt{y^2-1}} \textrm{d}y$$

By making the substitution $y=\sec{t}$, we may observe the following:

$$I = \int_0^x e^{-\sec^2t} \textrm{d}t$$

Now, I don't know much about this new integral, but I do know that $$I = \int_0^{\frac{\pi}{2}n} e^{-\sec^2t} \textrm{d}t = \dfrac{\pi}{2}n \left(1-\textrm{erf}(1)\right)$$ which gives me the feeling that the integral which you are trying to solve is non-elementary.

$\endgroup$
1
$\begingroup$

Firstly, $$I(z,c)=\int\limits_z^{\sqrt2} \dfrac{e^{-cy^2}\,\text dy}{y\sqrt{y^2-1}} =e^{-c}\int\limits_z^{\sqrt2}\dfrac{e^{-c(y^2-1)}}{y^2}\,\text d\sqrt{y^2-1} = e^{-c}\int\limits_{\sqrt{z^2-1}}^1 \dfrac{e^{-ct^2}\,\text dt}{t^2+1}.\tag1$$

At the second, are known the integrals $$\int\limits_0^\infty \dfrac{e^{-ct^2}\,\text dt}{t^2+1}\,\text dt = \dfrac\pi2 e^c\operatorname{erfc}\sqrt c,\tag2$$ $$\int\limits_0^1 \dfrac{e^{-ct^2}\,\text dt}{t^2+1}\,\text dt = \dfrac\pi4 e^c\left(1-\operatorname{erfc}^2\sqrt c\right),\tag3$$ with the closed form of $$I(1,c)=\dfrac\pi4\left(1-\operatorname{erfc}^2\sqrt c\right).\tag4$$

Besides, $$\int\limits_0^a \dfrac{t^{2k}\,\text dt}{t^2+1}\,\text dt = -\dfrac i2 (-1)^k \operatorname{B}\left(-a^2,k+\dfrac12,0\right),\tag5$$

The intermediate integral

$$\int\limits_0^a \dfrac{e^{-ct^2}\,\text dt}{t^2+1}\,\text dt = \sum\limits_{k=0}^\infty (-c)^k\int\limits_0^a \dfrac{t^{2k}\,\text dt}{t^2+1}\,\text dt =-\dfrac i2 \sum_{k=0}^\infty c^k\operatorname{B}\left(-a^2,k+\dfrac12,0\right),\tag6$$

$$I(z,c)=I(1,c)-e^{-c}\int\limits_0^{\sqrt{z^2-1}} \dfrac{e^{-ct^2}\,\text dt}{t^2+1}\,\text dt,\tag7$$ $$I(z,c)=I(1,c) + \dfrac i2 e^{-c} \sum_{k=0}^\infty c^k \operatorname{B}\left(1-z^2,k+\dfrac12,0\right).\tag8$$

Therefore, in the common case the given integral $(1)$ can be presented by the series $(8),$ which is not a closed form.

$\endgroup$
2
  • $\begingroup$ Thank you, Yuri! I gave the bounty to Omegadot, since the answer is easier to work with for me in the future. Nevertheless, I want to personally acknowledge and credit you for this nice result! $\endgroup$
    – drandran12
    Aug 21 '21 at 13:42
  • $\begingroup$ @drandran12 Thank you too. Of course, closed form is better. $\endgroup$ Aug 21 '21 at 14:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.