Binomial coefficients as multiple sum

April 5, 2020 / Leave a comment

Prove that

    \[\sum_{n_1=1}^{n-1} \sum_{n_2=1}^{n_1-1} \sum_{n_3=1}^{n_2-1} \cdots \sum_{n_m=1}^{n_{m-1}-1} 1 = \binom{n-1}{m}\]

Solution

The binomial coefficient in the RHS enumerates the subsets A of size m of \{1,2,\ldots,n-1\}. The LHS does the same thing, but choosing first the largest element n_1 of A, then its second-to-largest element n_2 < n_1, until choosing its smallest element n_m.

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Trigonometric integral

April 5, 2020 / Leave a comment

Prove that

    \[\bigintsss_0^{\infty}\frac{\sin \sqrt{x^2+1} \cos x}{\sqrt{x^2+1}} \, \mathrm{d}x=\frac{\pi}{4}\]

Solution

Let \mathcal{J} be the integral. Note that

    \[2 \sin \sqrt{x^2+1} \cos x = \sin \left ( \sqrt{1+x^2} - x \right ) + \sin \left ( \sqrt{1+x^2}+x \right ) \]

and hence:

    \[2\mathcal{J} = \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2}-x \right )}{\sqrt{1+x^2}} \,\mathrm{d}x + \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2} + x \right )}{\sqrt{1+x^2}} \, \mathrm{d}x\]

For the integral \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2}-x \right )}{\sqrt{1+x^2}} \,\mathrm{d}x we apply the substitution t \mapsto \sqrt{1+x^2}-x. Then, x = \frac{1-t^2}{2t} and

(1)   \begin{equation*} \bigintsss_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2}-x \right )}{\sqrt{1+x^2}} \,\mathrm{d}x = \int_{0}^{1} \frac{\sin t}{t} \, \mathrm{d}t \end{equation*}

and similarly by applying the change of variables t \mapsto \sqrt{1+x^2} + x at the second integral we get that

(2)   \begin{equation*} \int_{0}^{\infty} \frac{\sin \left ( \sqrt{1+x^2} \right )}{\sqrt{1+x^2}} \, \mathrm{d}x = \int_{1}^{\infty} \frac{\sin t}{t} \, \mathrm{d}t \end{equation*}

Adding equations (1) , (2) we get that

    \begin{align*} 2\mathcal{J} &= \int_{0}^{1} \frac{\sin t}{t} \, \mathrm{d}t + \int_{1}^{\infty} \frac{\sin t}{t} \, \mathrm{d}t \\ &= \int_{0}^{\infty} \frac{\sin t}{t} \, \mathrm{d}t \\ &= \frac{\pi}{2} \end{align*}

and the result follows.

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Parametric integral

April 3, 2020 / Leave a comment

Let a, b>0. Prove that:

    \[\int_{a}^{b} \frac{\mathrm{d}t}{t\sqrt{\left ( t-a \right )\left (b -t \right )} } = \frac{\pi}{\sqrt{ab}}\]

Solution

We’re applying the change of variables t \mapsto a \cos^2 \theta + b \sin^2 \theta and thus,

\begin{aligned} \int_{a}^{b} \frac{\mathrm{d}t}{t\sqrt{\left ( t-a \right )\left (b -t \right )}} &=\int_{0}^{\pi/2} \frac{2\left ( b-a \right )\sin \theta \cos \theta}{\left ( a \cos^2 \theta + b \sin^2 \theta \right ) \sqrt{\left ( b-a \right )^2 \sin^2 \theta \cos^2 \theta}} \, \mathrm{d}\theta \\ &=2 \int_{0}^{\pi/2} \frac{\mathrm{d} \theta}{a \cos^2 \theta + b \sin^2 \theta} \\ &=2 \int_{0}^{\pi/2} \frac{\mathrm{d}\theta}{\cos^2 \theta \left ( a + b \tan^2 \theta \right )} \\ &=2 \int_{0}^{\pi/2} \frac{\sec^2 \theta}{a + b \tan^2 \theta } \, \mathrm{d} \theta \\ &\!\!\!\!\!\overset{y = \tan \theta}{=\! =\! =\! =\! =\!} 2 \int_{0}^{\infty} \frac{\mathrm{d}y}{a + b y^2} \\ &= \frac{\pi}{\sqrt{ab}} \end{aligned}

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Abel’s Integral

April 2, 2020 / Leave a comment

Prove that:

    \[\int_{0}^{\infty} \frac{t}{\left ( e^{\pi t} -e^{-\pi t} \right )\left ( 1+t^2 \right )}\, \mathrm{d}t = \frac{\ln 2}{2} - \frac{1}{4}\]

Solution

We are applying Abel – Plana. We choose f(t)=\frac{1}{2(1+t)} thus,

\begin{aligned} i \int_{0}^{\infty} \frac{f\left ( it \right )- f\left ( -it \right )}{2 \sinh \pi t}\, \mathrm{d}t + \frac{f(0)}{2} = \sum_{n=0}^{\infty} (-1)^n f(n) &\Leftrightarrow i \int_{0}^{\infty} \frac{\frac{1}{2\left ( 1+it \right )}- \frac{1}{2\left ( 1-it \right )}}{2 \sinh \pi t}\, \mathrm{d}t + \frac{1}{4} = \\ &\quad \quad \quad \quad = \sum_{n=0}^{\infty} \frac{(-1)^n}{2\left ( 1+n \right )}\\ &\Leftrightarrow -i^2 \int_{0}^{\infty} \frac{t}{2\sinh \pi t \left ( t^2+1 \right )} \, \mathrm{d}t + \frac{1}{4} = \\ &\quad \quad \quad \quad = \sum_{n=0}^{\infty} \frac{(-1)^n}{2\left ( 1+n \right )} \\ &\Leftrightarrow \int_{0}^{\infty} \frac{t}{2\sinh \pi t \left ( t^2+1 \right )} \, \mathrm{d}t + \frac{1}{4} = \frac{\ln 2}{2} \\ &\Leftrightarrow \int_{0}^{\infty} \frac{t}{2\sinh \pi t \left ( t^2+1 \right )} \, \mathrm{d}t = \frac{\ln 2}{2} - \frac{1}{4} \\ &\Leftrightarrow \int_{0}^{\infty} \frac{t}{\left ( e^{\pi t} -e^{-\pi t} \right )\left ( 1+t^2 \right )}\, \mathrm{d}t = \frac{\ln 2}{2} - \frac{1}{4} \end{aligned}

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A convergent series

March 28, 2020 / Leave a comment

Let \sum \limits_{n=1}^{\infty} a_n be a convergent series of positive terms. Prove that there exists a strictly increasing sequence b_n which is also unbounded such that the series \sum \limits_{n=1}^{\infty} a_n b_n also converges.

Solution

We set b_n=\frac{1}{\sqrt{\sum \limits_{k=n}^{\infty}a_k}} and we observe that

    \[b_n a_n < 2\left(\sqrt{\sum_{k=n}^{\infty}a_k}- \sqrt{\sum_{k=n+1}^{\infty}a_k}\right)\]

That’s all folks.

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