25
$\begingroup$

Let $\operatorname{erfi}(x)$ be the imaginary error function $$\operatorname{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{z^2}dz.$$ Consider the integral $$I=\int_0^\infty\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt{x}\right)\ e^{-x\sqrt{2}}}{x}dx.$$ Its numeric value is approximately $0.625773669454426\dots$

Is it possible to express $I$ in a closed form using only elementary functions, integers and constants $\pi$, $e$?

$\endgroup$
  • 2
    $\begingroup$ Yes it is, and I am working on the derivation, but I get $$\tanh ^{-1}\left(\frac{\sqrt{2 \left(1-\sqrt{2}+\sqrt{4-2 \sqrt{2}}\right)}}{1+\sqrt{4-2 \sqrt{2}}}\right)$$ $\endgroup$ – Ron Gordon Sep 22 '13 at 1:45
  • 6
    $\begingroup$ @Marty $$\ln\sqrt{\frac1{\frac12-\sqrt{\frac1{6\sqrt{4-2\sqrt2}+6\sqrt2-2}}}-1}$$ $\endgroup$ – Vladimir Reshetnikov Sep 22 '13 at 2:13
  • $\begingroup$ @VladimirReshetnikov did you use some Computer Algebra System to get that value or did you get it by hand? If you get it by CAB, then please tell me how?? $\endgroup$ – Santosh Linkha Sep 22 '13 at 4:48
  • $\begingroup$ @experimentX I derived the result in a semi-manual way with some help from Mathematica. Unfortunately, neither Maple nor Mathematica is able to evaluate this integral directly. $\endgroup$ – Vladimir Reshetnikov Sep 24 '13 at 1:05
32
$\begingroup$

My strategy here is to use Parseval's equality to express the integral in a simpler form. This requires a strategic splitting of the integrand into Fourier transforms.

Begin by writing

$$\text{erfi}({\sqrt{x}})=\frac{2}{\sqrt{\pi}} \sqrt{x} \int_0^1 dt \, e^{x t^2}$$

and consider the following Fourier Transform:

$$\int_{-\infty}^{\infty} dx \, \theta(x) \, \text{erfi}(\sqrt{x}) \, e^{-\sqrt{2} x} \, e^{i k x}$$

where $\theta(x)$ is the Heaviside step function, which is $1$ when $x \gt 0$ and $0$ when $x \lt 0$. Using a change in the order of integration, we may evaluate this Fourier transform in exact form:

$$\begin{align}\int_{0}^{\infty} dx \, \text{erfi}(\sqrt{x}) \, e^{-\sqrt{2} x} \, e^{i k x} &=\frac{2}{\sqrt{\pi}} \int_0^1 dt \, \int_{0}^{\infty} dx \, \sqrt{x} e^{x t^2} \, e^{-\sqrt{2} x} \, e^{i k x}\\ &= \frac{2}{\sqrt{\pi}} \int_0^1 dt \, \int_{0}^{\infty} dx \, \sqrt{x} e^{-(\sqrt{2}-t^2-i k) x}\\ &= \int_0^1 \frac{dt}{(\sqrt{2}-i k - t^2)^{3/2}} \\ &=\frac{1}{\sqrt{2}-i k} \frac{1}{\sqrt{\sqrt{2}-1-i k}}\end{align} $$

Note that the third line comes from the integral

$$\int_0^{\infty} dx \, \sqrt{x} e^{-a x} = \frac{\sqrt{\pi}}{2 a^{3/2}}$$

The result in the fourth line may be obtained using a trig substitution in the integral in the third line; the only trick is pretending that $\sqrt{2}-i k$ may be set to some $b^2$ parameter, and then proceeding with the usual trig substitution.

Now, the rest of the original integrand is $\sin{x}/x$, which Fourier transform is simply $\pi$ when $|k| \lt 1$ and $0$ otherwise. We may then invoke Parseval's equality, which states that, for functions $f$ and $g$ and their respective Fourier transforms $F$ and $G$, we have

$$\int_{-\infty}^{\infty} dx \, f(x) g(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G(k)$$

Here,

$$f(x) = \text{erfi}(\sqrt{x}) \, e^{-\sqrt{2} x}\, \theta(x) $$ $$g(x) = \frac{\sin{x}}{x}$$ $$F(k) = \frac{1}{\sqrt{2}-i k} \frac{1}{\sqrt{\sqrt{2}-1-i k}}$$ $$G(k) = \begin{cases}\pi & |k| \lt 1 \\ 0 & k \gt 1\end{cases}$$

Thus, we have reduced the integral to the evaluation of the following:

$$\frac12 \int_{-1}^1 \frac{dk}{\sqrt{2}-i k} \frac{1}{\sqrt{\sqrt{2}-1-i k}}$$

Now sub $v^2=\sqrt{2}-1-i k$ and get the following integral

$$-i \int_{\sqrt{\sqrt{2}-1-i}}^{\sqrt{\sqrt{2}-1+i}} \frac{dv}{1+v^2}$$

which evaluates to

$$-i \left [\arctan{\sqrt{\sqrt{2}-1+i}} - \arctan{\sqrt{\sqrt{2}-1-i}} \right ] $$

which is equal to

$$\tanh ^{-1}\left(\frac{\sqrt{2 \left(1-\sqrt{2}+\sqrt{4-2 \sqrt{2}}\right)}}{1+\sqrt{4-2 \sqrt{2}}}\right) $$

or

$$\frac12 \log{\left [\frac{1+\sqrt{4-2 \sqrt{2}}+\sqrt{2 \left(1-\sqrt{2}+\sqrt{4-2 \sqrt{2}}\right)}}{1+\sqrt{4-2 \sqrt{2}}-\sqrt{2 \left(1-\sqrt{2}+\sqrt{4-2 \sqrt{2}}\right)}}\right ]} \approx 0.625774$$

$\endgroup$
  • 1
    $\begingroup$ @achillehui: you are too kind. You're not bad yourself, by the way. $\endgroup$ – Ron Gordon Sep 22 '13 at 4:35
  • $\begingroup$ @RonGordon This is great, thank you very much! $\endgroup$ – Marty Colos Sep 22 '13 at 18:41
  • $\begingroup$ Just a question: How much time did it take you to find out this solution? +1 btw. $\endgroup$ – user93957 Feb 7 '14 at 13:01
  • 1
    $\begingroup$ @Aðøbe: It's been long enough that I don't remember specifically, but it did take about a couple of hours to find the right path to the solution. $\endgroup$ – Ron Gordon Feb 7 '14 at 13:31
9
$\begingroup$

For $a>1$, $$\int_0^\infty\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt x\right)\ e^{-a x}}x dx=\ln\sqrt{\frac{\sqrt{a^2-2a+2}+\sqrt2\sqrt{\sqrt{a^2-2a+2}-a+1}+1}{\sqrt{a^2-2a+2}-\sqrt2\sqrt{\sqrt{a^2-2a+2}-a+1}+1}}.$$

$\endgroup$
  • 1
    $\begingroup$ +1. Is there an interpretation of the RHS when $a\leqslant1$? $\endgroup$ – Did Sep 22 '13 at 11:05
  • $\begingroup$ @did: see my derivation above. There are convergence problems for $a \le 1$. $\endgroup$ – Ron Gordon Sep 22 '13 at 11:10
  • 1
    $\begingroup$ This is precisely my point: the LHS does not exist when $a\lt1$ but the RHS does, for every $a$. $\endgroup$ – Did Sep 22 '13 at 11:15
  • 3
    $\begingroup$ @Did Yes, there is a possible interpretation. Let $s_n(a)=\displaystyle\int_0^{2\pi n}\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt x\right)\ e^{-a x}}x dx$ where $n\in\mathbb{N}$. The RHS gives the limit of the sequence $\lim\limits_{n\to\infty}s_n(a)$. Because $n$ increases discretely, in each wave of the integrand both half-waves can partially compensate each other, enabling the sequence to converge. I did not try to establish for which exactly values of $a$ the integral can be regularized this way. $\endgroup$ – Vladimir Reshetnikov Sep 22 '13 at 20:46
  • $\begingroup$ Nice. Thanks. $ $ $\endgroup$ – Did Sep 22 '13 at 20:53

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.