I want to prove the following relation: $${\Gamma_{1/2}}=\intop_{t=0}^{+\infty}\frac{e^{-t}}{\sqrt{t}}dt=2\sqrt{2}\intop_{x=0}^{+\infty}{\sin(x^2)}dx$$

I noticed that: $$\frac{\intop_{x=-\infty}^{+\infty}{e^{-x^2}}dt}{\sqrt{2}}=\intop_{x=-\infty}^{+\infty}{\sin(x^2)}dx=\intop_{x=-\infty}^{+\infty}{\cos(x^2)}dx=\frac{\sqrt{\pi}}{\sqrt{2}}$$

And consequently proved the relation by comparing the resulting value. However, I would like to know any alternate solutions that do not include the Gaussian integral. I have had no luck figuring this out, I would greatly appreciate any help.

  • $\begingroup$ See Fresnel integral $\endgroup$
    – Ricky
    Sep 7, 2023 at 16:26
  • $\begingroup$ Maybe this answers your question? $\endgroup$
    – Robert Lee
    Sep 7, 2023 at 19:08

1 Answer 1


I use the idea from Quanto in this post (Proof only by transformation that : $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $). Note $$ \frac1{\sqrt2}\int_0^\infty e^{-x^2}dx=\frac1{2\sqrt2}\int_0^\infty \frac{e^{-x}}{\sqrt x}dx, \int_0^\infty\sin(x^2)dx=\frac12\int_0^\infty\frac{\sin(x)}{\sqrt x}dx. $$ Let $$f(x)=\frac{\sqrt2}2e^{-x}-\sin x, g(x)=\frac1{\sqrt x}$$ and then $$ \mathcal{L}\{f\}(x)=\frac1{\sqrt2(x+1)}-\frac1{x^2+1}, \mathcal{L}\{g\}(x)=\sqrt{\frac\pi x}. $$ Now using $$ \int_0^\infty\mathcal{L} \{g\}(x)f(x)dx = \int_0^\infty g(x)\mathcal{L}\{f\}(x)dx $$ one has \begin{eqnarray} &&\frac1{\sqrt2}\int_0^\infty e^{-x^2}dx-\int_0^\infty\sin(x^2)dx\\ &=&\frac12\int_0^\infty f(x)g(x)dx=\frac1{2\sqrt\pi}\int_0^\infty f(x)\mathcal{L}\{g\}(x)dx\\ &=&\frac1{2\sqrt\pi}\int_0^\infty g(x)\mathcal{L}\{f\}(x)dx\\ &=&\frac1{2\sqrt\pi}\int_0^\infty\frac1{\sqrt x}\bigg(\frac1{\sqrt2(x+1)}-\frac1{x^2+1}\bigg)dx\\ &=&0. \end{eqnarray} So $$ \frac1{\sqrt2}\int_0^\infty e^{-x^2}dx=\int_0^\infty\sin(x^2)dx. $$


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