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Does anybody know how to perform the integral $$ \int_{-\infty}^\infty\frac{e^{-x^2}\sin(x) }{x } $$ Thanks.

Arthur
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user86406
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    Be sure to use $\LaTeX$. Also, be sure to show some thoughts for the problem. Thanks! – NasuSama Jul 15 '13 at 19:08
  • Ok, thank you. Current, my only idea is to Taylor expand the $\sin(x)/x$ and collect terms, but I'm very sure there are some better ideas. For example, if some body knows the integral representation of $\sin(x)/x$, it may be a useful method. – user86406 Jul 15 '13 at 19:13
  • See http://math.stackexchange.com/questions/208250/integral-of-sinc-function-multiplied-by-gaussian – Ross B. Jul 15 '13 at 19:21
  • I give my own answer to this problem. $1/x=\int_{0}^{\infty}e^{-xt}dt$, and write $\sin(x)$ using Euler formula. – user86406 Jul 15 '13 at 20:55

2 Answers2

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Write $$ \frac{\sin x}{x} = \int_{-1}^1 \frac12 e^{i x y}\,dy, $$ so that the integral is $$ \frac12\int_{-\infty}^\infty dx \int_{-1}^1\,dy \,e^{-x^2+i x y} = \frac12\int_{-1}^{1}\sqrt\pi e^{-y^2/4}\,dy = \frac\pi2 \mathop{\text{erf}}(x/2)\Big|_{-1}^{1} = \pi \mathop{\text{erf}}(1/2). $$

Kirill
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Consider : $$f(a)=\int_{-\infty}^\infty \frac{e^{-x^2}\sin(ax)}x\,dx$$ then \begin{align} f'(a)&=2\int_0^\infty e^{-x^2}\cos(ax)\,dx\\ &=\int_{-\infty}^\infty e^{-x^2}e^{iax}\,dx\\ &=\int_{-\infty}^\infty e^{-(x-ia/2)^2-a^2/4}\,dx\\ &=e^{-a^2/4}\int_{-\infty}^\infty e^{-(x-ia/2)^2}\,dx\\ &=e^{-a^2/4}\int_{-\infty}^\infty e^{-y^2}\,dy\quad(*)\\ &=\sqrt{\pi}\;e^{-a^2/4} \end{align}

($(*)$ a justification for this rewriting is provided in the comments)

Integrating again we get :

\begin{align} f(a)&=\sqrt{\pi}\int_0^a\,e^{-a^2/4}\,da\\ f(a)&=\pi\;\operatorname{erf}\left(\frac a2\right) \end{align}

(from the definition of the error function)

and the answer will be : $$f(1)=\pi\;\operatorname{erf}\left(\frac 12\right)$$

Raymond Manzoni
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  • I always think the parametrization trick is beautiful yet in a wicked way. +1. – Shuhao Cao Jul 15 '13 at 19:25
  • Thanks @Shuhao ! I enjoy the more general formulas obtained this way. – Raymond Manzoni Jul 15 '13 at 19:36
  • @Raymond. So your answer shows me that not only integral representation, but also derivative representation is useful to perform integrals. Thank you. – user86406 Jul 15 '13 at 19:42
  • Two remarks (other than this answer is incorrect). The fifth equality in the derivation of $f'(a)$ is not a simple variable substitution but requires some argument involving contour integration. Your "integrating again" step to obtain $f(a)$ misses an integration constant, which is in fact not zero. – WimC Jul 15 '13 at 19:44
  • @Wimc: If you mean the integration path parallel to the real line (of offset $-ia/2$) I knew that it didn't change the result (from playing with Fresnel integrals) but had not the time to insist on this. The $f(a)$ integration should be right since we are integrating $\operatorname{erf}$ that goes from $0$ to $a$ (with change of variable $t:=a/2$ so that the factor of $2$ disappears with the $2$ from erf definition). I had a missing factor of $2$ in the earlier part I think... – Raymond Manzoni Jul 15 '13 at 19:56
  • You are right. Sorry for the noise. +1. – WimC Jul 15 '13 at 20:39
  • No problem @WimC and thanks. – Raymond Manzoni Jul 15 '13 at 20:41
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    @user86406: The point of WimC was that, to make the derivation more rigorous, one should write something like $$\int_{-R}^R e^{-(x-ia/2)^2},dx=\int_{-R-ia/2}^{R-ia/2}e^{-y^2},dy=\\int_{-R-ia/2}^{-R} e^{-y^2},dy+\int_{-R}^R e^{-y^2},dy+\int_{R}^{R-ia/2} e^{-y^2},dy$$ Since $,y\mapsto e^{-y^2}$ is an entire function (so that the integral is not path dependent) we may take the limit as $R\to\infty$ to eliminate the two supplementary integrals on both sides of $\displaystyle\int_{-R}^R e^{-y^2},dy$. – Raymond Manzoni Jul 15 '13 at 21:24
  • The method exposed here is what is named 'differentiating under the integral sign' and was popularized by Richard Feynman see too Keith Conrad's paper. You will notice that another non rigorous point in this derivation (as well as many others using this method) is that we should justify the exchange of the derivation and integration operations – Raymond Manzoni Jul 15 '13 at 21:24
  • (this is often done using uniform convergence and the existence and convergence of the different terms). I am fond of this method but Kirill's method is nice too! – Raymond Manzoni Jul 15 '13 at 21:25