Zhengwen Liu

(2) $\begin{align*} A_{11} \,&=\, -{{\langle{1 2}\rangle}^4 [4 5]^3 \over s_{345} {\langle{1 6}\rangle} {\langle{2 6}\rangle} [3 4] \langle{1}| 4{+}5 |3] \langle{2}| 3{+}4 |5]} \\[0.5em] %% A_{12} \,&=\, -{{\langle{2 5}\rangle} \langle{2}| 5{+}6 |4]^3 \over {\langle{2 6}\rangle} {\langle{5 6}\rangle} [3 4] {\langle{2}| 3{+}4 |1]} {\langle{5}| 3{+}4 |1]} {\langle{5}| 2{+}6 |3]}} \\[0.5em] %% A_{13} \,&=\, -{[3 6]\, {\langle{1}| 4{+}5 |6]}^3 \over {\langle{4 5}\rangle} [2 6]\, [2 3]\, {\langle{1}| 4{+}5 |3]} {\langle{4}| 2{+}3 |6]} {\langle{5}| 2{+}6 |3]}} \\[0.5em] %% A_{14} \,&=\, {{\langle{2 3}\rangle}^3 [5 6]^3 {\langle{2}| 3{+}4 |6]} \over s_{561} {\langle{3 4}\rangle} [1 6]\, {\langle{2}| 3{+}4 |1]} {\langle{2}| 3{+}4 |5]} {\langle{4}| 2{+}3 |6]}} \\[0.5em] %% A_{15} \,&=\, {{\langle{3}| 4{+}5 |6]}^3 \over s_{345} {\langle{3 4}\rangle} {\langle{4 5}\rangle} [1 2]\, [2 6]\, {\langle{5}| 3{+}4 |1]}} \end{align*}$

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Integral via Inversions

Posted on March 21, 2019 by zhengwen

Let us compute the following integral

(1) $\begin{align*} \int_0^{\infty} {\ln (2x) \over 1+x^2}dx \end{align*}$

Performing $x\to {1\over x}$ , this integral becomes

(2) $\begin{align*} \int_{\infty}^{0} {\ln (2x^{-1}) \over 1+x^{-2}}\bigg(-\frac{dx}{x^2}\bigg) \,=\, \int_0^{\infty} {\ln (2x^{-1}) \over 1+x^2} dx \end{align*}$

Thus we have

(3) $\begin{align*} \int_0^{\infty} {\ln (2x) \over 1+x^2}dx \,&=\, {1\over 2} \int_0^{\infty} {\ln (2x) + \ln (2x^{-1}) \over 1+x^2}dx \\ \,&=\, {\ln 2} \int_0^{\infty} {dx \over 1+x^2} \\ \,&=\, {\pi \ln 2 \over 2} \end{align*}$

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Fresnel integral

Posted on March 20, 2019 by zhengwen

The Fresnel integrals read

(1) $\begin{align*} {\cal S} \,=\, \int_0^{\infty}dt\, \sin t^2 \\ {\cal C} \,=\, \int_0^{\infty}dt\, \cos t^2 \end{align*}$

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Consider the following integral

(2) $\begin{align*} {\sqrt{\pi} \over 2} \,=\, \int_0^{\infty}dz\,e^{-z^2} \,&=\, -\int_{\gamma_3}dz\,e^{-z^2} \\ \,&=\, e^{i\pi/4} \int_0^{\infty}dt\,e^{-it^2} \\ \,&=\, {1+i \over \sqrt{2}} \int_0^{\infty}dt\,\big(\cos t^2 - i\sin t^2\big) \\ \,&=\, {1 \over \sqrt{2}} \int_0^{\infty}dt\,\Big( \big(\cos t^2 + \sin t^2\big) +i \big(\cos t^2 - \sin t^2\big)\Big) \\ \,&=\, {1 \over \sqrt{2}} \Big( \big({\cal C} + {\cal S}\big) +i \big({\cal C} - {\cal S}\big)\Big) \end{align*}$

From this we have

(3) $\begin{align*} {\cal C} + {\cal S} \,&=\, {\sqrt{\pi} \over \sqrt{2}} \\ {\cal C} - {\cal S} \,&=\, 0 \end{align*}$

which gives

(4) $\begin{align*} {\cal C} \,=\, {\cal S} \,=\, {\sqrt{\pi} \over \sqrt{8}} \end{align*}$

Equation (2) also gives

(5) $\begin{align*} \int_{-\infty}^{\infty}dt\,e^{-it^2} \,=\, e^{-i\pi/4} \int_{-\infty}^{\infty}dz\,e^{-z^2} \,=\, \sqrt{\pi} e^{-i\pi/4} \end{align*}$

We give an alternative derivation in the following. Let us start with

(6) $\begin{align*} G(x) \,=\, \bigg(\int_0^{x}dt\, e^{it^2}\bigg)^2 + i\int_0^1 dt\, {e^{ix^2(t^2+1)} \over t^2 + 1} \end{align*}$

Differentiating $G(x)$ with respect to $x$

(7) $\begin{align*} {dG(x) \over dx} \,&=\, 2\bigg(\int_0^{x}dt\, e^{it^2}\bigg)e^{ix^2} + i\int_0^1 dt\, {i2x(t^2+1) e^{ix^2(t^2+1)} \over t^2 + 1} \\ \,&=\, 2\bigg(\int_0^{x}dt\, e^{it^2}\bigg)e^{ix^2} -2x \int_0^1 dt\, e^{ix^2(t^2+1)} \\ \,&=\, 2\bigg(\int_0^{x}dt\, e^{it^2}\bigg)e^{ix^2} -2e^{ix^2}\int_0^x du\, e^{iu^2} \,=\, 0 \end{align*}$

where we chnage the variable, $u=xt$ .
Therefore we have

(8) $\begin{align*} G(x) \,\equiv\, G(0) \,=\, i\int_0^1 dt {1 \over t^2 + 1} \,=\, i{\pi \over 4} \end{align*}$

Since in the limit $x\to \infty$

(9) $\begin{align*} \lim_{x\to\infty}\int_0^1 dt\, {e^{ix^2(t^2+1)} \over t^2 + 1} \,=\, 0 \end{align*}$

we have

(10) $\begin{align*} G(\infty) \,=\, \bigg(\int_0^{\infty}dt\, e^{it^2}\bigg)^2 \,=\, \bigg({\cal C} + i {\cal S}\bigg)^2 \end{align*}$

Solving the equation

(11) $\begin{align*} i{\pi \over 4} \,=\, {\cal C}^2 - {\cal S}^2 + 2i\, {\cal C}{\cal S} \end{align*}$

gives (4).

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Integral by Euler’s Formula

Posted on March 20, 2019 by zhengwen

We compute the following integral

(1) $\begin{align*} \int_{1}^{\infty}{dx \over x\, (x^2 + 1)} \end{align*}$

Performing a partial fraction expansion of the integrand

(2) $\begin{align*} {1 \over x\, (x^2 + 1)} \,=\, {1 \over x\, (x + i)(x - i)} \,=\, {1 \over x} - {1 \over 2(x + i)} - {1 \over 2(x - i)} \end{align*}$

Thus we have

(3) $\begin{align*} \int_{1}^{\infty}dx\, \bigg( {1 \over x} - {1 \over 2(x + i)} - {1 \over 2(x - i)} \bigg) \,&=\, \Big(\ln x - 2\ln(x + i) - 2\ln(x - i)\Big)\Big|_{1}^{\infty} \\[0.5em] \,&=\, \bigg(\ln{x \over \sqrt{x^2 + 1}}\bigg)\bigg|_{1}^{\infty} \\[1.1em] \,&=\, \ln\sqrt{2} \end{align*}$

Let us see another integral

(4) $\begin{align*} \int_{0}^{\infty}dx\, \sin(\alpha x)\, e^{-\beta x} \end{align*}$

Using Euler’s Formula

(5) $\begin{align*} \sin(\alpha x) \,=\, {e^{i\alpha x} - e^{-i\alpha x} \over 2i} \end{align*}$

we have

(6) $\begin{align*} \int_{0}^{\infty}dx\, \sin(\alpha x)\, e^{-\beta x} \,&=\, {1 \over 2i}\int_{0}^{\infty}dx\, \Big(e^{(i\alpha -\beta) x} - e^{-(i\alpha + \beta) x}\Big) \\ \,&=\, {1 \over 2i}\bigg({e^{(i\alpha -\beta)x} \over i\alpha -\beta} - {e^{-(i\alpha + \beta) x} \over -(i\alpha + \beta)}\Big)\bigg|_{0}^{\infty} \\ \,&=\, {1 \over 2i}\bigg({1 \over \beta - i\alpha} - {1 \over \beta + i\alpha}\Big) \\ \,&=\, {\alpha \over \alpha^2 + \beta^2} \end{align*}$

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Gaussian integral

Posted on March 20, 2019 by zhengwen

The Gaussian integral is:
\begin{align}
\int_{-\infty}^{\infty}dx e^{-x^2} \,=\, \sqrt{\pi}
\end{align}
A standard way to derive this formula is
\begin{align}
\bigg(\int_{-\infty}^{\infty}dx e^{-x^2}\bigg)^2
\,&=\, \int_{-\infty}^{\infty}dx \int_{-\infty}^{\infty}dy\, e^{-(x^2+y^2)}
\\
\,&=\, \int_0^{2\pi} d\phi \int_{0}^{\infty}dr\,r e^{-r^2}
\\
\,&=\, \int_0^{2\pi} d\phi \int_{0}^{\infty}{dr^2 \over 2}\, e^{-r^2}
\\
\,&=\, \pi
\end{align}
Performing the change of variable $x=\sqrt{t}$, this turns into the Euler integral
\begin{align}
\int_{-\infty}^{\infty}dx\, e^{-x^2} \,=\, 2 \int_0^\infty dx\, e^{-x^2}
\,=\, 2\int_0^\infty dt\, \frac{1}{2}\ e^{-t} \ t^{-\frac{1}{2}}
\,=\, \Gamma\left(\tfrac{1}{2}\right) = \sqrt{\pi}
\end{align}
where $\Gamma$ is the gamma function. From this we can see that the factorial of a half-integer is a rational multiple of $\sqrt{\pi}$. More generally, let us consider
\begin{align}
\int_0^{\infty} dx\, e^{-\alpha x^\beta}
\end{align}
Change the varible $t=\alpha x^\beta$,
\begin{align}
dx \,=\, {1 \over \alpha^{1\over \beta} \beta}\, t^{{1\over\beta} – 1}\, dt
\end{align}
and thus we have
\begin{align}
\int_0^{\infty} dx\, e^{-\alpha x^\beta}
\,=\, {1 \over \alpha^{1\over \beta} \beta}\,\int_0^{\infty} dt\, t^{{1\over\beta} – 1}\,e^{-t} dt
\,=\, {\Gamma\left(\tfrac{1}{\beta}\right) \over \alpha^{1\over \beta} \beta}
\end{align}

Another type of integrals is
\begin{align}
2\int_0^{\infty} dx\,x^{2n} e^{-x^2}
\,=\, \int_0^{\infty} dt\,t^{n-{1\over 2}} e^{-t}
\,=\, \Gamma(n+\tfrac{1}{2})
\end{align}

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Hello world!

Posted on February 13, 2019 by zhengwen

Welcome to WordPress. This is your first post. Edit or delete it, then start writing!

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