I am facing issue to evaluate this particular type of definite integral,

$$I=\int_{0}^{\infty}\exp(-\alpha x^2)\cos(\beta x)dx$$

Please suggest a way to this.



closed as off-topic by Carl Mummert, Ethan Bolker, Silvia Ghinassi, Pragabhava, Watson Oct 24 '16 at 16:29

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    $\begingroup$ This is just a Fourier transform of $e^{-\alpha x^2}$ $\endgroup$ – Boby Oct 22 '16 at 15:22
  • $\begingroup$ This is similar to Fourier transform. For a gaussian, the F.T. is also a gaussian. $\endgroup$ – Srini Oct 22 '16 at 15:24
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    $\begingroup$ Why has this question been put on hold? It has sparked a number of good solutions which will undoubtedly help many viewers. A question like this is what makes MSE interesting, informative, and enjoyable. $\endgroup$ – poweierstrass Oct 26 '16 at 15:24

Let $I(\alpha,\beta)$ be given by the integral

$$I(\alpha,\beta)=\int_0^\infty e^{-\alpha x^2}\cos(\beta x)\,dx$$

Using Euler's Formula, $\cos(\beta x)=\text{Re}(e^{i\beta x})$. Then, we can write

$$\begin{align} I(\alpha,\beta)&=\text{Re}\left(\int_0^\infty e^{-\alpha x^2}e^{i\beta x}\,dx\right)\\\\ &=\text{Re}\left(\int_0^\infty e^{-\alpha (x^2-i(\beta/\alpha) x)}\,dx\right)\tag 1 \\\\ &=e^{-\beta^2/4\alpha}\text{Re}\left(\int_0^\infty e^{-\alpha (x-i(\beta/2\alpha) )^2}\,dx\right)\tag 2\\\\ &=e^{-\beta^2/4\alpha}\text{Re}\left(\int_{-i(\beta/2\alpha)}^{\infty-i(\beta/2\alpha)} e^{-\alpha x^2}\,dx\right) \tag 3\\\\ &=e^{-\beta^2/4\alpha}\text{Re}\left(\int_0^\infty e^{-\alpha x^2}\,dx\right) \tag 4\\\\ &=\sqrt{\frac{\pi}{4\alpha}}e^{-\beta^2/4\alpha} \end{align}$$


In going from $(1)$ to $(2)$, we completed the square in the exponent.

In going from $(2)$ to $(3)$, we enforced the substitution $x\to x+i\beta/2\alpha$.

In going from $(3)$ to $(4)$, we exploited Cauchy's Integral Theorem to deform the contour back to the real line. The real part operation nullifies the contribution from the integral from $-i\beta/2\alpha$ to $0$.

  • $\begingroup$ But the Euler's formula, I know is $\cos(\beta x) = \frac{\exp(i\beta x)+\exp(-i\beta x)}{2}$. How come $\cos(\beta x)=\text{Re}(e^{i\beta x})$? $\endgroup$ – zhk Oct 22 '16 at 15:48
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    $\begingroup$ @mmm Yes, you are correct. And $e^{i\beta x}=\cos(\beta x)+i\sin(\beta x)$ from which we find $\cos(\beta x)=\text{Re}(e^{i\beta x})$. $\endgroup$ – Mark Viola Oct 22 '16 at 15:52
  • $\begingroup$ If I am correct the (3) to (4) is to use $\oint\limits_{C}f(z)\,\mathrm{d}z=\int_{R}f(z)dz$? $\endgroup$ – zhk Oct 22 '16 at 16:15
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    $\begingroup$ The integrand is entire, so $\oint_C e^{-\alpha z^2}\,dz=0$. Thus, with $C$ the "rectangular" contour with vertices $0$, $\infty$, $\infty-i\beta/2\alpha$, and $-i\beta/2\alpha$, we have $$\int_{0}^\infty e^{-\alpha x^2}\,dx+\int_{\infty}^{\infty -i\beta/2\alpha}e^{-\alpha z^2}\,dz+\int_{\infty -i\beta/2\alpha}^{-i\beta/2\alpha}e^{-\alpha z^2}\,dz+\int_{-i\beta/2\alpha}^0 e^{-\alpha z^2}\,dz=0$$The second integral is zero since $e^{-t^2}\to 0$ as $t\to \infty$. The fourth integral is purely imaginary since $dz=idy$ and vanishes upon taking the real part. $\endgroup$ – Mark Viola Oct 22 '16 at 16:20
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    $\begingroup$ You're welcome. My pleasure. -Mark $\endgroup$ – Mark Viola Oct 22 '16 at 16:33

$$ \begin{align} \int_0^\infty e^{-\alpha x^2}e^{i\beta x}\,\mathrm{d}x &=\int_0^\infty e^{-\alpha\left(x-i\frac\beta{2\alpha}\right)^2-\frac{\beta^2}{4\alpha}}\mathrm{d}x\tag{1}\\ &=e^{-\frac{\beta^2}{4\alpha}}\int_{-i\frac\beta{2\alpha}}^{\infty-i\frac\beta{2\alpha}}e^{-\alpha x^2}\mathrm{d}x\tag{2}\\ &=e^{-\frac{\beta^2}{4\alpha}}\left[\int_{-i\frac\beta{2\alpha}}^0e^{-\alpha x^2}\mathrm{d}x+\int_0^\infty e^{-\alpha x^2}\mathrm{d}x+\color{#A0A0A0}{\int_\infty^{\infty-i\frac\beta{2\alpha}} e^{-\alpha x^2}\mathrm{d}x}\right]\tag{3}\\ &=e^{-\frac{\beta^2}{4\alpha}}\left[\,\color{#D080F8}{i\int_0^{\frac\beta{2\alpha}}e^{\alpha x^2}\mathrm{d}x}+\frac12\sqrt{\frac\pi\alpha}\,\right]\tag{4} \end{align} $$ Explanation:
$(1)$: complete the square
$(2)$: substituted $x\mapsto x+i\frac\beta{2\alpha}$
$(3)$: the integrand is entire; apply Cauchy's Integral Theorem
$(4)$: the integral over $\left[R,R-i\frac\beta{2\alpha}\right]$ vanishes

The integral in purple is pure imaginary, so taking the real parts of $(4)$ gives $$ \int_0^\infty e^{-\alpha x^2}\cos(\beta x)\,\mathrm{d}x =\frac12e^{-\frac{\beta^2}{4\alpha}}\sqrt{\frac\pi\alpha}\tag{5} $$

  • $\begingroup$ See the answer I wrote, in my opinion the identity theorem is easier to understand than the Cauchy integral theorem when you don't know so much of complex analysis $\endgroup$ – reuns Oct 22 '16 at 22:59

\begin{equation} I = \int\limits_{0}^{\infty} \mathrm{e}^{-ax^{2}} \cos(bx) \mathrm{d}x \tag{1} \label{eq:1} \end{equation}

Let \begin{align} I_{1} &= \int\limits_{0}^{\infty} \mathrm{e}^{-ax^{2}} \mathrm{e}^{ibx} \mathrm{d}x = \int\limits_{0}^{\infty} \mathrm{e}^{-ax^{2}+ibx} \mathrm{d}x \\ &= \mathrm{e}^{-b^{2}/4a} \int\limits_{0}^{\infty} \mathrm{e}^{-a(x-ib/2a)^{2}} \mathrm{d}x \\ \tag{2} \label{eq:2} \end{align} Here we completed the square and note that $\mathrm{Re}\,I_{1} = I$.

Consider the indefinite integral \begin{align} \int \mathrm{e}^{-a(x-ib/2a)^{2}} \mathrm{d}x &= \int \mathrm{e}^{-ay^{2}} \mathrm{d}y \\ &= \frac{1}{\sqrt{a}} \int \mathrm{e}^{-z^{2}} \mathrm{d}z = \frac{1}{2} \sqrt{\frac{\pi}{a}} \mathrm{erf}(z) \\ &= \frac{1}{2} \sqrt{\frac{\pi}{a}} \mathrm{erf}\left(x\sqrt{a} - \frac{ib}{2\sqrt{a}}\right) \tag{3} \label{eq:3} \end{align} we used the substitutions, $y=x- \frac{ib}{2a}$ and $z^{2} = ay^{2}$.

Examining the error function expression, we have \begin{equation} \lim_{x \to 0} \mathrm{erf}\left(x\sqrt{a} - \frac{ib}{2\sqrt{a}}\right) = \mathrm{erf}\left(- \frac{ib}{2\sqrt{a}}\right) = -i\,\mathrm{erfi}\left(\frac{b}{2\sqrt{a}}\right) \tag{4} \label{eq:4} \end{equation} which is a pure imaginary quantity with the assumption that all of the variables in the argument of the imaginary error function are real and $a \gt 0$. We also have \begin{equation} \lim_{x \to \infty} \mathrm{erf}\left(x\sqrt{a} - \frac{ib}{2\sqrt{a}}\right) = 1 \tag{5} \label{eq:5} \end{equation}

Using equations \eqref{eq:4} and \eqref{eq:5} in equation \eqref{eq:3} we obtain \begin{equation} \mathrm{Re}\left(\int\limits_{0}^{\infty} \mathrm{e}^{-a(x-ib/2a)^{2}} \mathrm{d}x \right) = \frac{1}{2} \sqrt{\frac{\pi}{a}} \tag{6} \label{eq:6} \end{equation}

Combining equations \eqref{eq:6} and \eqref{eq:2} yields our final result \begin{equation} \int\limits_{0}^{\infty} \mathrm{e}^{-ax^{2}} \cos(bx) \mathrm{d}x = \frac{1}{2} \sqrt{\frac{\pi}{a}} \,\mathrm{e}^{-b^{2}/4a} \end{equation}


The least amount of complex analysis is as follow :

  • For $z > 0$, let $$f(z) = \int_{-\infty}^\infty e^{-x^2} e^{2zx}dx$$ Complete the square $x^2-2zx= (x-z)^2 - z^2$ so that $$f(z) = e^{-z^2} \int_{-\infty}^\infty e^{-(x-z)^2}dx \underset{y = x-z}= e^{-z^2} \int_{-\infty}^\infty e^{-y^2}dy = Ce^{-z^2}$$ (where $C = \sqrt{\pi}$)

Then note that for every $z \in \mathbb{C}$ : $\int_{-\infty}^\infty e^{-x^2} e^{2zx}dx$ is analytic in $z$, as well as $Ce^{-z^2}$, thus by the identity theorem for analytic functions $$\forall z \in \mathbb{C}, \qquad f(z) = Ce^{-z^2}$$

  • Apply the same trick to $$g_z(s) = \int_{-\infty}^\infty e^{-sx^2} e^{2zx}dx$$ where $s > 0$. With $y = s^{1/2}x$ we get $$g_z(s) = s^{-1/2}\int_{-\infty}^\infty e^{-y^2} e^{2zs^{-1/2}y}dy = s^{-1/2}f(zs^{-1/2}) = \frac{C}{s^{1/2}}e^{-z^2/s}$$ Finally note that $\int_{-\infty}^\infty e^{-sx^2} e^{2zx}dx$ is analytic in $s$ for every $s \in \mathbb{C},Re(s) > 0$, as well as $\frac{C}{s^{1/2}}e^{-z^2/s}$, thus by the identity theorem $$\forall s \in \mathbb{C},Re(s) > 0, \qquad g_z(s) = \frac{C}{s^{1/2}}e^{-z^2/s}$$

  • Hence for any $Re(a) > 0$ : $$I = \frac{1}{2}\int_{-\infty}^\infty e^{-a x^2} \cos( \beta x)dx = \frac{g_{i \beta/2}(a)+g_{-i \beta/2}(a)}{4} = \frac{C}{2a^{1/2}}e^{\textstyle- \frac{\beta^2}{4a}}$$

  • $\begingroup$ How do you justify $$e^{-z^2} \int_{-\infty}^\infty e^{-(x-z)^2}dx \underset{y = x-z}= e^{-z^2} \int_{-\infty}^\infty e^{-y^2}dy$$ I believe you need Cauchy's Integral Theorem for this. $\endgroup$ – robjohn Oct 23 '16 at 0:27
  • $\begingroup$ @robjohn no read again, I proved it for $z > 0$ (I meant $z \in \mathbb{R}$ a typo) $\endgroup$ – reuns Oct 23 '16 at 0:32
  • $\begingroup$ You've shown that $f(z)=Ce^{-z^2}$ for $z\in\mathbb{R}$. How do you apply the identity theorem without showing it for $z$ in a non-empty, open subset of $\mathbb{C}$? $\endgroup$ – robjohn Oct 23 '16 at 1:29
  • $\begingroup$ @robjohn see en.wikipedia.org/wiki/Identity_theorem#An_improvement $\endgroup$ – reuns Oct 23 '16 at 1:34
  • $\begingroup$ @robjohn the idea is that $f(z) = g(z)$ for $z \in [a,b]$ means $f^{(k)}(a) = g^{(k)}(a)$ for every $k \in \mathbb{N}$, so they agree on an open around $a$. Otherwise, use that non-constant analytic functions have isolated zeros. $\endgroup$ – reuns Oct 23 '16 at 1:36

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