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 $\pi$, $e$?

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    $\begingroup$ 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, 2013 at 1:45
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    $\begingroup$ @Marty $$\ln\sqrt{\frac1{\frac12-\sqrt{\frac1{6\sqrt{4-2\sqrt2}+6\sqrt2-2}}}-1}$$ $\endgroup$ Sep 22, 2013 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?? $\endgroup$
    – S L
    Sep 22, 2013 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. $\endgroup$ Sep 24, 2013 at 1:05

2 Answers 2


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

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


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

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

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


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

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    $\begingroup$ @achillehui: you are too kind. You're not bad yourself, by the way. $\endgroup$
    – Ron Gordon
    Sep 22, 2013 at 4:35
  • $\begingroup$ @RonGordon This is great, thank you very much! $\endgroup$ Sep 22, 2013 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, 2014 at 13:01
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    $\begingroup$ @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, 2014 at 13:31

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

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    $\begingroup$ +1. Is there an interpretation of the RHS when $a\leqslant1$? $\endgroup$
    – Did
    Sep 22, 2013 at 11:05
  • $\begingroup$ @did: see my derivation above. There are convergence problems for $a \le 1$. $\endgroup$
    – Ron Gordon
    Sep 22, 2013 at 11:10
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    $\begingroup$ This is precisely my point: the LHS does not exist when $a\lt1$ but the RHS does, for every $a$. $\endgroup$
    – Did
    Sep 22, 2013 at 11:15
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    $\begingroup$ @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. $\endgroup$ Sep 22, 2013 at 20:46
  • $\begingroup$ Nice. Thanks. $ $ $\endgroup$
    – Did
    Sep 22, 2013 at 20:53

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