Generalized Fresnel Integrals

The generalized fresnel integrals, also known as Böhmer’s integrals, appear in Volume 2 of Higher Transcendental Functions (Bateman Manuscript), Section 9.10, Equations 1 and 2:

\mathrm{C}(x,a) &= \int\limits_{x}^{\infty} t^{a-1} \cos(t) \mathrm{d}t \\
&= \int\limits_{x}^{\infty} t^{a-1}\, \frac{\mathrm{e}^{it}+\mathrm{e}^{-it}}{2} \mathrm{d}t \\
&= \frac{1}{2} \Big[\mathrm{e}^{i\frac{\pi}{2}a} \Gamma(a,-ix) + \mathrm{e}^{-i\frac{\pi}{2}a} \Gamma(a,ix) \Big]

\mathrm{S}(x,a) &= \int\limits_{x}^{\infty} t^{a-1} \sin(t) \mathrm{d}t \\
&= \int\limits_{x}^{\infty} t^{a-1}\, \frac{\mathrm{e}^{it}-\mathrm{e}^{-it}}{i2} \mathrm{d}t \\
&= \frac{1}{i2} \Big[\mathrm{e}^{i\frac{\pi}{2}a} \Gamma(a,-ix) – \mathrm{e}^{-i\frac{\pi}{2}a} \Gamma(a,ix) \Big]

To derive these results, we require a result from Volume 1 of Higher Transcendental Functions (Bateman Manuscript), Section 1.5.1:

Next consider \(\int_{C} z^{a-1} \mathrm{e}^{cz} \mathrm{d}z\) where the contour C consists of the real axis from \(+\epsilon\) to +R, the arc of the circle \(z=R\mathrm{e}^{i\phi}\) from \(\phi = 0\) to \(\phi = \beta \,\, (-\pi /2 \leq \beta \leq \pi /2)\), the straight line from \(z=R\mathrm{e}^{i\beta}\) to \(\epsilon \mathrm{e}^{i\beta}\), and the arc of the circle \(z=\epsilon \mathrm{e}^{i\phi}\) from \(\phi = \beta\) to \(\phi = 0\). Since the value of the contour integral is zero, on making \(\epsilon \to 0\) and \(R \to \infty\) it follows that
\int\limits_{0}^{\infty} t^{a-1} \mathrm{e}^{-ct\cos(\beta)\,-ict\sin(\beta)} \mathrm{d}t
= \Gamma(a) c^{-a} \mathrm{e}^{-ia\beta}
-\frac{1}{2}\pi \lt \beta \lt \frac{1}{2}\pi ,\,\,\,\, \Re a \gt 0,\,\, \mathrm{or} \,\,\, \beta = \pm \frac{1}{2}\pi ,\,\,\,\, 0 \lt \Re a \lt 1

We begin our derivation with
\int\limits_{x}^{\infty} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t
= \int\limits_{0}^{\infty} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t \,- \int\limits_{0}^{x} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t
for the first integral in equation \eqref{eq:gfi-4}, via the reference above, for \(\beta = \pi /2\) and \(c = 1\), equation \eqref{eq:gfi-3} becomes
\int\limits_{0}^{\infty} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t = \Gamma(a) \mathrm{e}^{-i\frac{\pi}{2}a}
for the second integral in equation \eqref{eq:gfi-4}, we use the substitution \(y=it\)
\int\limits_{0}^{x} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t = \frac{1}{i^{a}} \int\limits_{0}^{ix} y^{a-1}\, \mathrm{e}^{-y} \mathrm{d}y = \mathrm{e}^{-i\frac{\pi}{2}a} \gamma(a,ix)

Substituting equations \eqref{eq:gfi-5} and \eqref{eq:gfi-6} into equation \eqref{eq:gfi-4} yields
\int\limits_{x}^{\infty} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t = \mathrm{e}^{-i\frac{\pi}{2}a} [\Gamma(a) \,-\, \gamma(a,ix)] = \mathrm{e}^{-i\frac{\pi}{2}a} \Gamma(a,ix)

Using the same arguments, beginning with substituting \(\beta = -\pi /2\) into equation \eqref{eq:gfi-3} we have
\int\limits_{x}^{\infty} t^{a-1}\, \mathrm{e}^{it} \mathrm{d}t = \mathrm{e}^{i\frac{\pi}{2}a} [\Gamma(a) \,-\, \gamma(a,-ix)] = \mathrm{e}^{i\frac{\pi}{2}a} \Gamma(a,-ix)

Substituting equations \eqref{eq:gfi-7} and \eqref{eq:gfi-8} into equations \eqref{eq:gfi-1b} and \eqref{eq:gfi-2b} yields the generalized fresnel integrals.


  1. \(\gamma(a,x) = \int_{0}^{x} t^{a-1} \mathrm{e}^{-t} \mathrm{d}t \) is the lower incomplete gamma function.
  2. \(\Gamma(a,x) = \int_{x}^{\infty} t^{a-1} \mathrm{e}^{-t} \mathrm{d}t \) is the upper incomplete gamma function.
  3. \(\Gamma(a) = \Gamma(a,x) + \gamma(a,x)\)
  4. This analysis was necessary in order to avoid difficulties when using the upper incomplete gamma function directly with the left hand sides of equations \eqref{eq:gfi-4} and \eqref{eq:gfi-8}. Doing so results, after a change in variables, in an upper limit of integration of \(i\infty\). If one treats this as a modulus and replaces it with \(\infty\) one obtains the correct results for the generalized fresnel integrals. However, I have not been able to justify this, thus the method presented here.

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