5
$\begingroup$

I have tried to evaluate the integral $$\int_{0}^\infty \frac{\sin(x)}{x}e^{- x^2} dx.$$ I used integration by parts but I did not succeed. Wolfram Alpha says that is convergent and it is equal to: $\frac \pi 2 \text{erf}({\frac 12})$ .

Is there any simple way for evaluate it?

$\endgroup$
  • $\begingroup$ Do it via the convolution theorem for the Fourier transform. $\endgroup$ – Cameron Williams Feb 9 '18 at 18:37
  • $\begingroup$ A simplified version of Cameron Williams’ suggestion is that you can consider $$\frac{\sin x}{x}=\frac{1}{2}\int_{-1}^{1}e^{ixt}\,dt$$ and then change the order of integration. $\endgroup$ – Sangchul Lee Feb 9 '18 at 18:46
6
$\begingroup$

We have that $$\begin{align}\int_{0}^\infty \frac{\sin(x)}{x}e^{-x^2} dx &= \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}\int_{0}^\infty x^{2k}e^{-x^2} dx\\ &= \frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}\int_{0}^\infty t^{k-1/2}e^{-t} dt\\ &=\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k\Gamma(k+1/2)}{(2k+1)!} \\ &=\sqrt{\pi}\sum_{k=0}^{\infty}\frac{(-1)^k\cdot(1/2)^{2k+1}}{(2k+1)k!}= \frac{\pi}{2}\text{erf}(1/2).\end{align}$$ For the last step see the Taylor series of erf.

$\endgroup$
2
$\begingroup$

Here is a slightly longer method (compared to @Robert Z's slick series solution approach) that uses Feynman's trick of differentiating under the integral sign.

Let $$I(a) = \int_0^\infty \frac{\sin (ax)}{x} e^{-x^2} \, dx, \quad a > 0.$$ Note that $I(0) = 0$ and we are required to find $I(1)$.

Differentiating $I(a)$ with respect to the parameter $a$ gives $$I'(a) = \int_0^\infty \cos (ax) e^{-x^2} \, dx. \tag1$$

On integrating (1) by parts leads to $$I'(a) = \frac{2}{a} \int_0^\infty x e^{-x^2} \sin (ax) \, dx.$$

Also, differentiating (1) again with respect to the parameter $a$ yields $$I''(a) = -\int_0^\infty x e^{-x^2} \sin (ax) \, dx = -\frac{a}{2} I'(a).$$ If we set $u(a) = I'(a)$ the above second-order differential equation can be reduced to the following first-order differential equation $$u'(a) = -\frac{a}{2} u(a).$$ Solving yields $$u(a) = I'(a) = K e^{-a^2/4}, \tag1$$ where $K$ is a constant to be determined. To find this constant setting $a = 0$ in $I'(a)$ leads to $$I'(0) = \int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \cdot \frac{2}{\sqrt{\pi}} \int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \cdot \text{erf} (\infty) = \frac{\sqrt{\pi}}{2}.$$ So on setting $a= 0$ in (1) we find $K = \sqrt{\pi}/2$. Thus $$I'(a) = \frac{\sqrt{\pi}}{2} e^{-a^2/4}.$$

Now as $I(0) = 0$, we have $$I(1) = \int_0^1 I'(a) \, da = \frac{\sqrt{\pi}}{2} \int_0^1 e^{-a^2/4} \, da.$$ Enforcing a substitution of $a \mapsto 2a$ leads to $$I(1) = \sqrt{\pi} \int_0^{1/2} e^{-a^2} \, da = \frac{\pi}{2} \cdot \frac{2}{\sqrt{\pi}} \int_0^{1/2} e^{-a^2} \, da.$$ And since from the integral representation for the error function which is given by $$\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt,$$ one has $$\int_0^\infty \frac{\sin x}{x} e^{-x^2} \, dx = \frac{\pi}{2} \text{erf} \left (\frac{1}{2} \right ),$$ as expected.

$\endgroup$
1
$\begingroup$

Following Sangchul Lee's hint. \begin{align} I:=\int^\infty_0 \frac{\sin(x)}{x}e^{-x^2}\,dx=\frac{1}{2}\int^\infty_{-\infty}\frac{\sin(x)}{x}e^{-x^2}\,dx=\frac{1}{4}\int^\infty_{-\infty}\int^1_{-1}e^{ixt}e^{-x^2}\,dt\,dx \end{align} Since $|e^{ixt}e^{-x^2}|=e^{-x^2}$ the double integral is clearly absolutely convergent so by Tonelli-Fubini we can interchange the integration order to obtain: \begin{align} I=\frac{1}{4}\int^{1}_{-1}\int^{\infty}_{-\infty}e^{ixt}e^{-x^2}\,dx\,dt=\frac{1}{4}\int^{1}_{-1}\int^{\infty}_{-\infty}e^{-\frac{t^2}{4}-(x-it/2)^2}\,dx\,dt \end{align} where we have completed the square on the last integral. Now by an easy contour integration on a rectangle we get \begin{align} \int^\infty_{-\infty} e^{ixt}e^{-x^2}\,dx=\sqrt[]{\pi}e^{-t^2/4} \end{align} So with substitutation and parity (P) we can conclude: \begin{align} I=\frac{\sqrt[]{\pi}}{4}\int^1_{-1}e^{-t^2/4}\,dt\stackrel{\color{red}{u=t/2}}{=}\frac{\sqrt[]{\pi}}{2}\int^{1/2}_{-1/2}e^{-u^2}\,du\stackrel{\color{red}{P}}{=}\frac{\pi}{2}\cdot\frac{2}{\sqrt[]{\pi}}\int^{1/2}_0e^{-u^2}\,du=\frac{\pi}{2}\operatorname{erf}\left(\frac 12\right) \end{align}

$\endgroup$

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.