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Integrals and sequence

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Papapetros Vaggelis
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Integrals and sequence

#1

Post by Papapetros Vaggelis » Thu Jul 14, 2016 9:02 am

\(\displaystyle{1)}\) Calculate the integral \(\displaystyle{I=\int_{-\infty}^{0}\dfrac{2\,x^2-1}{x^4+1}\,\mathrm{d}x}\) .

\(\displaystyle{2)}\) Calculate the integral \(\displaystyle{J=\int_{-\infty}^{\infty}\dfrac{\sin\,2\,x}{x^2-2\,x+5}\,\mathrm{d}x}\) .

\(\displaystyle{3)}\) Prove that the real sequence \[\displaystyle{a_{n}=1+r\,\cos\,a+r^2\,\cos\,2\,a+...+ r^{n}\,\cos\,(n\,a)\,,n\in\mathbb{N}}\] converges and find the limit \(\displaystyle{\lim_{n\to \infty}a_{n}}\) , where \(\displaystyle{r\in\left(0,1\right)}\) and \(\displaystyle{a\in\left(0,\frac{\pi}{2}\right)}\) .
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Tolaso J Kos
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Re: Integrals and sequence

#2

Post by Tolaso J Kos » Thu Jul 14, 2016 9:04 am

Good evening Vaggelis.
I'll answer the second one for now.
We want to evaluate the integral \( \displaystyle \int_{-\infty}^{\infty}\frac{\sin 2x}{x^2-2x+5}\,dx \), which clearly converges (absolutely) because: $$\int_{-\infty}^{\infty}\left|\frac{\sin 2x}{x^2-2x+5} \right| \,dx\leq \int_{-\infty}^{\infty}\frac{dx}{x^2-2x+5}=\frac{\pi}{2}$$ thus the integral converges.

Now, consider the function \( \displaystyle f(z)=\frac{e^{2iz}}{z^2-2z+5} \) which is analytic on the complex plane and has (simple) poles \( \displaystyle z_1=1+2i, \; z_2=1-2i \). Only \(z_1 \) lies on the upper half plane. The residue at \(z_1 \) is: $$\begin{aligned}
\mathfrak{Res}\left ( f, z_1 \right ) &=\lim_{z\rightarrow z_1}\left ( z-z_1 \right )f(z) \\
&=\lim_{z\rightarrow z_1}\left ( z-z_1 \right )\frac{e^{2iz}}{\left ( z-z_1 \right )\left ( z-z_2 \right )} \\
&= \lim_{z\rightarrow z_1}\frac{e^{2iz}}{z-z_2}\\
&= \frac{e^{2i\left ( 1+2i \right )}}{1+2i-1+2i}\\
&= \frac{\sin 2}{4e^4}-\frac{i}{4e^4}
\end{aligned}$$ Thus the integral is evaluated to: $$\int_{-\infty}^{\infty}\frac{\sin 2x}{x^2-2x+5}\,dx=2\pi \mathfrak{Re}\left ( \mathfrak{Res}\left ( f, z_1 \right ) \right )=\frac{2\pi\sin 2}{4e^4}=\frac{\pi \sin 2}{2e^4}$$

T:
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Tsakanikas Nickos
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Re: Integrals and sequence

#3

Post by Tsakanikas Nickos » Thu Jul 14, 2016 9:08 am

Good evening to both of you! Let me answer (3).

It is known that \[ \displaystyle \left( \cos\theta + i\sin\theta \right)^{n} = \cos(n\theta) + i\sin(n\theta) \] Let \[ \displaystyle S_{n} = \sum_{k=0}^{n} \left[ r \left( \cos a + i\sin a \right) \right]^{k} \] and observe that \( \displaystyle Re(S_{n}) = a_{n} \, , \, n \in \mathbb{N}. \) We have that \begin{align*}
S_{n} &= \sum_{k=0}^{n} \left[ r \left( \cos a + i\sin a \right) \right]^{k} \\ \\
&= \frac{ \left[ r \left( \cos a + i\sin a \right) \right]^{n+1} - 1 }{ r \left( \cos a + i\sin a \right) - 1} \\ \\
&= \frac{ \left\{ \left[ r \left( \cos a + i\sin a \right) \right]^{n+1} - 1 \right\} \left\{ ( r\cos a - 1) - ir\sin a \right\} }{ (r\cos a -1)^2 + (r\sin a)^2 } \\ \\
&= \frac{ \left\{ \left[ r \left( \cos a + i\sin a \right) \right]^{n+1} - 1 \right\} \left\{ ( r\cos a - 1) - ir\sin a \right\} }{r^2 -2r\cos a +1 } \\ \\
&= \frac{ \left\{ r^{n+1} \left[ \; \cos((n+1)a) +i\sin((n+1)a) \; \right] -1 \right\} \left\{ ( r\cos a - 1) - ir\sin a \right\} }{r^2 -2r\cos a +1 }
\end{align*} and now it is easy to see that \begin{align*}
Re(S_{n}) &= \frac{1}{r^2 -2r\cos a +1} \left\{ \left[ r^{n+1}\cos( (n+1)a ) - 1 \right] \left[ r\cos a -1 \right] + \left[ r^{n+1}\sin( (n+1)a ) \right] \left[ r\sin a \right] \right\} \\ \\
&= \frac{1}{r^2 -2r\cos a +1} \left\{ r^{n+2}\cos a \cos( (n+1)a ) - r^{n+1}\cos( (n+1)a ) - r\cos a +1 + r^{n+2}\sin a \sin( (n+1)a ) \right\} \\ \\
&= \frac{1}{r^2 -2r\cos a +1} \left\{ r^{n+2}\cos(na) - r^{n+1}\cos( (n+1)a ) - r\cos a + 1 \right\}
\end{align*}
(I hope that these computations are correct!)

Since \( r \in (0,1) \) , \( \alpha \in ( 0 , \frac{\pi}{2} ) \) and \( \cos\theta \) is a bounded function, it follows that \[ \displaystyle \lim_{n} a_{n} = \lim_{n} Re(S_{n}) = \frac{ 1 - r\cos a }{ r^2 -2r\cos a +1 } \]
Papapetros Vaggelis
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Re: Integrals and sequence

#4

Post by Papapetros Vaggelis » Thu Jul 14, 2016 9:10 am

Thank you for your answers.

Here is a solution for the first integral.

Let \(\displaystyle{x<0}\) and \(\displaystyle{I(x)=\int_{x}^{0}\dfrac{2\,t^2-1}{t^4+1}\,\mathrm{d}t}\) .

The function \(\displaystyle{t\mapsto \dfrac{2\,t^2-1}{t^4+1}\,,t\in\left[x,0\right]}\) is continuous at \(\displaystyle{\left[x,0\right]}\)

and thus \(\displaystyle{I(x)\in\mathbb{R}}\) .

\(\displaystyle{t\in\mathbb{R}\implies t^4+1=\left(t^2+1\right)^2-2\,t^2=\left(t^2-\sqrt{2}\,t+1\right)\,\left(t^2+\sqrt{2}\,t+1\right)}\) .

Let \(\displaystyle{A\,,B\,,C\,,D\in\mathbb{R}}\) such that :

\(\displaystyle{\dfrac{2\,t^2-1}{t^4+1}=\dfrac{A\,t+B}{t^2-\sqrt{2}\,t+1}+\dfrac{C\,t+D}{t^2+\sqrt{2}\,t+1}\,\forall\,t\in\mathbb{R}}\)

or equivalently,

\(\displaystyle{2\,t^2-1=\left(A\,t+B\right)\,\left(t^2+\sqrt{2}\,t+1\right)+\left(C\,t+D\right)\,\left(t^2-\sqrt{2}\,t+1\right)\,\forall\,t\in\mathbb{R}}\)

or :

\(\displaystyle{2\,t^2-1=\left(A+C\right)\,t^3+\left(\sqrt{2}\,A+B-\sqrt{2}\,C+D\right)\,t^2+\left(A+\sqrt{2}\,B+C-\sqrt{2}\,D\right)\,t+\left(B+D\right)\,\forall\,t\in\mathbb{R}}\) .

From the last equation, we get :

\(\displaystyle{\left(A+C,\sqrt{2}\,A+B-\sqrt{2}\,C+D,A+\sqrt{2}\,B+C-D\,\sqrt{2},B+D\right)=\left(0,2,0,-1\right)}\)

so : \(\displaystyle{\left(A,B,C,D\right)=\left(\dfrac{3}{2\,\sqrt{2}},-\dfrac{1}{2},-\dfrac{3}{2\,\sqrt{2}},-\dfrac{1}{2}\right)}\) .

Let's verify :

\(\displaystyle{\begin{aligned}\dfrac{A\,t+B}{t^2-\sqrt{2}\,t+1}+\dfrac{C\,t+D}{t^2+\sqrt{2}\,t+1}&=\dfrac{1}{2\,\sqrt{2}}\,\left[\dfrac{3\,t-\sqrt{2}}{t^2-\sqrt{2}\,t+1}+\dfrac{-3\,t-\sqrt{2}}{t^2+\sqrt{2}\,t+1}\right]\\&=\dfrac{1}{2\,\sqrt{2}}\,\dfrac{\left(3\,t-\sqrt{2}\right)\,\left(t^2+\sqrt{2}\,t+1\right)+\left(-3\,t-\sqrt{2}\right)\,\left(t^2-\sqrt{2}\,t+1\right)}{\left(t^2-\sqrt{2}\,t+1\right)\,\left(t^2+\sqrt{2}\,t+1\right)}\\&=\dfrac{1}{2\,\sqrt{2}}\,\dfrac{4\,\sqrt{2}\,t^2-2\,\sqrt{2}}{t^4+1}\\&=\dfrac{2\,t^2-1}{t^4+1}\end{aligned}}\)

Therefore,

\(\displaystyle{\begin{aligned} I(x)&=\dfrac{1}{2\,\sqrt{2}}\,\int_{x}^{0}\left[\dfrac{3\,t-\sqrt{2}}{t^2-\sqrt{2}\,t+1}-\dfrac{3\,t+\sqrt{2}}{t^2+\sqrt{2}\,t+1}\right]\,\mathrm{d}t\\&=\dfrac{1}{4\,\sqrt{2}}\,\int_{x}^{0}\left[\dfrac{6\,t-2\,\sqrt{2}}{t^2-\sqrt{2}\,t+1}-\dfrac{6\,t+2\,\sqrt{2}}{t^2+\sqrt{2}\,t+1}\right]\,\mathrm{d}t\\&=\dfrac{1}{4\,\sqrt{2}}\,\int_{x}^{0}\left[\dfrac{3\,\left(2\,t-\sqrt{2}\right)}{t^2-\sqrt{2}\,t+1}+\dfrac{\sqrt{2}}{t^2-\sqrt{2}\,t+1}-\dfrac{3\,\left(2\,t+\sqrt{2}\right)}{t^2+\sqrt{2}\,t+1}+\dfrac{\sqrt{2}}{t^2+\sqrt{2}\,t+1}\right]\,\mathrm{d}t\\&=\int_{x}^{0}\left[\dfrac{3}{4\,\sqrt{2}}\,\left(\dfrac{2\,t-\sqrt{2}}{t^2-\sqrt{2}\,t+1}-\dfrac{2\,t+\sqrt{2}}{t^2+\sqrt{2}\,t+1}\right)+\dfrac{1}{2\,sqrt{2}}\,\left(\dfrac{\sqrt{2}}{1+\left(\sqrt{2}\,t+1\right)^2}+\dfrac{\sqrt{2}}{1+\left(\sqrt{2}\,t-1\right)^2}\right)\right]\,\mathrm{d}t\\&=\left[\dfrac{3}{4\,\sqrt{2}}\,\ln\,\dfrac{t^2-\sqrt{2}\,t+1}{t^2+\sqrt{2}\,t+1}+\dfrac{1}{2\,\sqrt{2}}\,\arctan\,\left(\sqrt{2}\,t+1\right)+\dfrac{1}{2\,\sqrt{2}}\,\arctan\,\left(\sqrt{2}\,t-1\right)\right]_{x}^{0}\\&=\dfrac{3}{4\,\sqrt{2}}\,\ln\,\dfrac{x^2+\sqrt{2}\,x+1}{x^2-\sqrt{2}\,x+1}-\dfrac{1}{2\,\sqrt{2}}\,\arctan\,\left(\sqrt{2}\,x+1\right)-\dfrac{1}{2\,\sqrt{2}}\,\arctan\,\left(\sqrt{2}\,x-1\right)\end{aligned}}\)

Since

\(\displaystyle{\lim_{x\to -\infty}\dfrac{x^2+\sqrt{2}\,x+1}{x^2-\sqrt{2}\,x+1}=1\,,\lim_{x\to -\infty}\left(\sqrt{2}\,x\pm 1\right)=-\infty}\) , we have that :

\(\displaystyle{\lim_{x\to -\infty}I(x)=0-\dfrac{1}{2\,\sqrt{2}}\,\left(-\dfrac{\pi}{2}\right)-\dfrac{1}{2\,\sqrt{2}}\,\left(-\dfrac{\pi}{2}\right)=\dfrac{\pi}{2\,\sqrt{2}}\in\mathbb{R}}\) .

So, \(\displaystyle{I=\int_{-\infty}^{0}\dfrac{2\,x^2-1}{x^4+1}\,\mathrm{d}x=\lim_{x\to -\infty}I(x)=\dfrac{\pi}{2\,\sqrt{2}}}\) .
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Tolaso J Kos
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Re: Integrals and sequence

#5

Post by Tolaso J Kos » Thu Jul 14, 2016 9:13 am

Papapetros Vaggelis wrote:\(\displaystyle{1)}\) Calculate the integral \(\displaystyle{I=\int_{-\infty}^{0}\dfrac{2\,x^2-1}{x^4+1}\,\mathrm{d}x}\) .
A solution for this one using contour integration over a counterclockwise semicircle contour on the upper half plane with center at the origin and radius \( R \).

Since the integrand is an even function we are evaluating the integral \( \displaystyle \mathcal{J}=\int_{-\infty}^{\infty}\frac{2x^2-1}{x^4+1}\,dx \).

For this purpose we consider the contour integral: \( \displaystyle \oint_{\gamma}\frac{2z^2-1}{z^4+1}\,dz \). We note that the function \( \displaystyle f(z)=\frac{2z^2-1}{z^4+1} \) has four simple poles which only two lie in the contour. Those two poles are:
$$z_1=\frac{1+i}{\sqrt{2}}, \;\; z_2=\frac{-1+i}{\sqrt{2}}$$ and the residues at those poles are \( \displaystyle \mathfrak{Res}\left ( f;z_1 \right )=\lim_{z\rightarrow z_1}\left ( z-z_1 \right )f(z)=\cdots=\frac{3\sqrt{2}}{8}-i\frac{\sqrt{2}}{8}\) and \( \displaystyle \mathfrak{Res}\left ( f;z_2 \right )=\lim_{z\rightarrow z_2}\left ( z-z_2 \right )f(z)=\cdots=-\frac{3\sqrt{2}}{8}-i\frac{\sqrt{2}}{8} \).

It still holds that: $$\oint_{\gamma}f(z)=2\pi i \sum res=2\pi i \left ( \frac{3\sqrt{2}}{8}-i\frac{\sqrt{2}}{8} -\frac{3\sqrt{2}}{8}-i\frac{\sqrt{2}}{8} \right )=\frac{\pi}{\sqrt{2}}$$
Expanding the contour integral we have:
$$\oint_{\gamma}f(z)\,dz=\int_{-R}^{R}+\int_{{\rm arc}}$$ Letting \(R \to +\infty\) it is easy to note/prove that the arc integral goes to zero. Therefore:
$$\int_{-\infty}^{\infty}f(x)\,dx=\frac{\pi}{\sqrt{2}}$$ Hence the original integral is the half of that, meaning that: \( \displaystyle \int_{-\infty}^{0}f(x)\,dx=\frac{\pi}{2\sqrt{2}} \).
Imagination is much more important than knowledge.
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Re: Integrals and sequence

#6

Post by Tolaso J Kos » Thu Jul 14, 2016 9:20 am

Papapetros Vaggelis wrote:\(\displaystyle{2)}\) Calculate the integral \(\displaystyle{J=\int_{-\infty}^{\infty}\dfrac{\sin\,2\,x}{x^2-2\,x+5}\,\mathrm{d}x}\) .
Similarly we are applying contour integration over a semicircle contour on the upper half plane with radius \( R \).
The function \( \displaystyle f(z)=\frac{e^{2iz}}{z^2-2z+5} \) has two simple poles , but only one of them is encircled in the contour.
The residue at that pole is:
$$\mathfrak{Res}\left ( f;1+2i \right )=\lim_{z\rightarrow z_1}\left ( z-z_1 \right )f(z)=\cdots=\frac{\sin 2}{4e^4}-\frac{i}{4e^4}$$

The contour integral is equal to:
$$\oint_{\gamma}f(z)\,dz=2\pi i \mathfrak{Res}\left ( 1+2i \right )=2\pi i \left ( \frac{\sin 2}{4e^4}-\frac{i}{4e^4} \right )$$

Expanding the contour we have that:
$$\oint_{\gamma}f(z)\,dz=\int_{-R}^{R}+\int_{{\rm arc}}$$
Letting \( R \rightarrow +\infty\) that arc integral goes to zero. So:
$$\int_{-\infty}^{\infty}\frac{\sin 2x}{x^2-2x+5}\,dx=\mathfrak{Im}\left ( \oint_{\gamma}f(z)\,dz \right )=\frac{\pi \sin 2}{2e^4}$$

Note: If we were to take the real part we would have taken the value of the integral: \( \displaystyle \int_{-\infty}^{\infty}\frac{\cos 2x}{x^2-2x+5}\,dx \).
Imagination is much more important than knowledge.
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