Unlikely application of Parseval’s equality

A poster somehow got a numerical value for an integral and wondered if it could be expressed in terms of simple constants.

Let $\operatorname{erfi}(x)$ be the imaginary error function
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 $0.625773669454426$

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

Of course it is – otherwise, why would I post it here? Anyway, here is my solution…

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 Foureir 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 evaluated 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

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


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

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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