- DESY Welcome Services

International Office (Residence Permit & Insurance)

Registration in Hamburg

My WordPress Blog

- DESY Welcome Services

International Office (Residence Permit & Insurance)

Registration in Hamburg

Posted in Uncategorized
Leave a comment

Let us compute the following integral

(1)

Performing , this integral becomes

(2)

Thus we have

(3)

Posted in Uncategorized
Leave a comment

The Fresnel integrals read

(1)

Consider the following integral

(2)

From this we have

(3)

(4)

Equation (2) also gives

(5)

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

(6)

Differentiating with respect to

(7)

where we chnage the variable, .

Therefore we have

(8)

Since in the limit

(9)

we have

(10)

Solving the equation

(11)

gives (4).

Posted in Uncategorized
Leave a comment

We compute the following integral

(1)

Performing a partial fraction expansion of the integrand

(2)

Thus we have

(3)

Let us see another integral

(4)

Using Euler’s Formula

(5)

we have

(6)

Posted in Uncategorized
Leave a comment

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}

Posted in Uncategorized
Leave a comment

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

Posted in Uncategorized
Leave a comment