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- Jan 17, 2013

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\(\displaystyle \int^{\infty}_{0} \frac{\sin (x) }{ x} = \frac{\pi }{2}\)

There are three different methods to solve the integral .

- Thread starter ZaidAlyafey
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- Jan 17, 2013

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\(\displaystyle \int^{\infty}_{0} \frac{\sin (x) }{ x} = \frac{\pi }{2}\)

There are three different methods to solve the integral .

- Feb 13, 2012

- 1,704

$\displaystyle \mathcal{L} \{f(t)\} = F(s) \implies \mathcal{L} \{\frac{f(t)}{t}\} = \int_{s}^{\infty} F(u)\ du$ (1)

... so that is...

$\displaystyle \mathcal{L} \{\sin t\} = \frac{1}{1 + s^{2}} \implies \mathcal{L} \{ \frac{\sin t}{t}\} = \int_{s}^{\infty} \frac {d u}{1+ u^{2}} = \frac{\pi}{2} - \tan^{-1} s$ (2)

... and computing (2) for s=0 we find...

$\displaystyle \int_{0}^{\infty} \frac{\sin t}{t}\ dt = \frac{\pi}{2}$ (3)

Kind regards

$\chi$ $\sigma$

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- #3

- Jan 17, 2013

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Excellent . Still two other ways to solve it .

$\displaystyle \mathcal{L} \{f(t)\} = F(s) \implies \mathcal{L} \{\frac{f(t)}{t}\} = \int_{s}^{\infty} F(u)\ du$ (1)

... so that is...

$\displaystyle \mathcal{L} \{\sin t\} = \frac{1}{1 + s^{2}} \implies \mathcal{L} \{ \frac{\sin t}{t}\} = \int_{s}^{\infty} \frac {d u}{1+ u^{2}} = \frac{\pi}{2} - \tan^{-1} s$ (2)

... and computing (2) for s=0 we find...

$\displaystyle \int_{0}^{\infty} \frac{\sin t}{t}\ dt = \frac{\pi}{2}$ (3)

Kind regards

$\chi$ $\sigma$

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- #4

- Mar 5, 2012

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$$\mathcal{F} \{\text{sinc }x\}(\xi) = \text{rect } \xi$$

$$\int_{-\infty}^{+\infty} \frac {\sin(\pi x)}{\pi x} e^{-2\pi i \xi x}dx = \text{rect } \xi$$

For $\xi = 0$ this is:

$$\int_{-\infty}^{+\infty} \frac {\sin(\pi x)}{\pi x} dx = 1$$

Substituting $\pi x = t$ gives:

$$\int_{-\infty}^{+\infty} \frac {\sin t}{t} dx = \pi$$

And since $\frac {\sin t}{t}$ is symmetric, we get:

$$\int_{0}^{\infty} \frac {\sin t}{t} dx = \frac \pi 2$$

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- #5

- Jan 17, 2013

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I didn't know it can be solved this way . I had two other methods in mind

$$\mathcal{F} \{\text{sinc }x\}(\xi) = \text{rect } \xi$$

$$\int_{-\infty}^{+\infty} \frac {\sin(\pi x)}{\pi x} e^{-2\pi i \xi x}dx = \text{rect } \xi$$

For $\xi = 0$ this is:

$$\int_{-\infty}^{+\infty} \frac {\sin(\pi x)}{\pi x} dx = 1$$

Substituting $\pi x = t$ gives:

$$\int_{-\infty}^{+\infty} \frac {\sin t}{t} dx = \pi$$

And since $\frac {\sin t}{t}$ is symmetric, we get:

$$\int_{0}^{\infty} \frac {\sin t}{t} dx = \frac \pi 2$$

- Feb 13, 2012

- 1,704

A milestone of complex analysis is the use of Cauchy Integral Theorems for solving integrals that are 'resistent' to 'conventional' attaks. In particular the first of these theorems extablishes that if a complex variable function is analytic inside and along a closed path C is...

$\displaystyle \int_{C} f(z)\ d z = 0$ (1)

Now we computing (1) in the case $\displaystyle f(z)= \frac{e^{i\ z}}{z}$ when the path is represented in the figure...

... and when $R \rightarrow \infty$ and $r \rightarrow 0$. The Jordan's lemma extablishes that the integral along the 'big circle' tends to 0 if $ R \rightarrow \infty$ , so that, tacking into account (1), we conclude that is...

$\displaystyle \int_{- \infty}^{+ \infty} \frac{\sin x}{x}\ dx = \lim_{r \rightarrow 0} \int_{0}^{\pi} e^{i\ r\ e^{i \theta}}\ d \theta = \pi$ (2)

Details have been omitted but it's clear that this approach is more complex and less elegant that the Laplace Transform approach...

Kind regards

$\chi$ $\sigma$

$\displaystyle \int_{C} f(z)\ d z = 0$ (1)

Now we computing (1) in the case $\displaystyle f(z)= \frac{e^{i\ z}}{z}$ when the path is represented in the figure...

... and when $R \rightarrow \infty$ and $r \rightarrow 0$. The Jordan's lemma extablishes that the integral along the 'big circle' tends to 0 if $ R \rightarrow \infty$ , so that, tacking into account (1), we conclude that is...

$\displaystyle \int_{- \infty}^{+ \infty} \frac{\sin x}{x}\ dx = \lim_{r \rightarrow 0} \int_{0}^{\pi} e^{i\ r\ e^{i \theta}}\ d \theta = \pi$ (2)

Details have been omitted but it's clear that this approach is more complex and less elegant that the Laplace Transform approach...

Kind regards

$\chi$ $\sigma$

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- Jan 17, 2013

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A complex analysis approach is necessarily complex , but I don't I agree it is less elegant .Details are been omitted but it's clear that this approach is more complex and less elegant that the Laplace Transform approach...

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- #8

- Jan 17, 2013

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The last method

See hint

See hint

Differentiation under the integral sign

- Jan 31, 2012

- 253

You can also use contour integration to evaluate $\displaystyle \int_{0}^{\infty} \frac{\sin^{n} x}{x^{n}} \ dx \ (n = 2, 3, \ldots) $.

$\displaystyle \int_{0}^{\infty} \frac{\sin^{n} x}{x^{n}} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{n} x}{x^{n}} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{1}{x^{n}} \Big( \frac{e^{ix}-e^{-ix}}{2i} \Big)^{n} \ dx$

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{1}{(x-i \epsilon)^{n}} \Big( e^{ix}-e^{-ix} \Big)^{n} \ dx $

The last line is justified for $n >1$ by the dominated convergence theorem.

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{1}{(x-i \epsilon)^{n}} \sum_{k=0}^{n} \binom{n}{k} (e^{ix})^{n-k} (-e^{-ix})^{k} \ dx $ (binomial theorem)

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{1}{(x-i \epsilon)^{n}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} e^{ix(n-2k)} \ dx $

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \int_{-\infty}^{\infty} \frac{e^{ix(n-2k)}}{(x- i \epsilon)^{n}} \ dx$

Now let $\displaystyle f(z) = \frac{e^{iz(n-2k)}}{(z- i \epsilon)^{n}}$ and integrate around a large closed half-circle in the upper-half complex plane if $n-2k \ge 0$, and around a large closed half-circle in the lower-half plane if $n - 2k < 0$.

There will be no contribution from the integrals around the half-circle in the lower-half plane since the pole is in the upper half plane. So we'll be summing from $k=0$ to the largest integer such that $k \le \frac{n}{2}$.

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} \ 2 \pi i \ \text{Res}[f,i \epsilon]$

The pole at $z=i \epsilon$ is of order $n$.

So $\displaystyle \text{Res} [f, i \epsilon] = \frac{1}{(n-1)!} \lim_{z \to i \epsilon} \frac{d^{n-1}}{dx^{n-1}} \ e^{iz(n-2k)} = \frac{1}{(n-1)!} \lim_{z \to i \epsilon} i^{n-1} (n-2k)^{n-1} e^{iz(n-k)} $

$ \displaystyle = \frac{1}{(n-1)!} i^{n-1} (n-2k)^{n-1} e^{-\epsilon(n-k)} $

And $\displaystyle \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} 2 \pi i \ \text{Res}[f,i \epsilon] $

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} 2 \pi i \ \frac{1}{(n-1)!} i^{n-1} (n-2k)^{n-1} e^{-\epsilon(n-k)}$

$ \displaystyle = \frac{\pi}{2^{n}(n-1)!} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} (n-2k)^{n-1} $

So, for example, $\displaystyle \int_{0}^{\infty} \frac{\sin^{5} x}{x^{5}} \ dx = \frac{\pi}{2^{5}4!} \sum_{k=0}^{2} \binom{5}{k} (-1)^{k}(5-2k)^{4}= \frac{\pi}{768} \Big(5^4-5(3)^{4} +10(1)^4 \Big) = \frac{115 \pi}{384}$

$\displaystyle \int_{0}^{\infty} \frac{\sin^{n} x}{x^{n}} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{n} x}{x^{n}} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{1}{x^{n}} \Big( \frac{e^{ix}-e^{-ix}}{2i} \Big)^{n} \ dx$

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{1}{(x-i \epsilon)^{n}} \Big( e^{ix}-e^{-ix} \Big)^{n} \ dx $

The last line is justified for $n >1$ by the dominated convergence theorem.

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{1}{(x-i \epsilon)^{n}} \sum_{k=0}^{n} \binom{n}{k} (e^{ix})^{n-k} (-e^{-ix})^{k} \ dx $ (binomial theorem)

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{1}{(x-i \epsilon)^{n}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} e^{ix(n-2k)} \ dx $

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \int_{-\infty}^{\infty} \frac{e^{ix(n-2k)}}{(x- i \epsilon)^{n}} \ dx$

Now let $\displaystyle f(z) = \frac{e^{iz(n-2k)}}{(z- i \epsilon)^{n}}$ and integrate around a large closed half-circle in the upper-half complex plane if $n-2k \ge 0$, and around a large closed half-circle in the lower-half plane if $n - 2k < 0$.

There will be no contribution from the integrals around the half-circle in the lower-half plane since the pole is in the upper half plane. So we'll be summing from $k=0$ to the largest integer such that $k \le \frac{n}{2}$.

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} \ 2 \pi i \ \text{Res}[f,i \epsilon]$

The pole at $z=i \epsilon$ is of order $n$.

So $\displaystyle \text{Res} [f, i \epsilon] = \frac{1}{(n-1)!} \lim_{z \to i \epsilon} \frac{d^{n-1}}{dx^{n-1}} \ e^{iz(n-2k)} = \frac{1}{(n-1)!} \lim_{z \to i \epsilon} i^{n-1} (n-2k)^{n-1} e^{iz(n-k)} $

$ \displaystyle = \frac{1}{(n-1)!} i^{n-1} (n-2k)^{n-1} e^{-\epsilon(n-k)} $

And $\displaystyle \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} 2 \pi i \ \text{Res}[f,i \epsilon] $

$ \displaystyle = \frac{1}{2} \frac{1}{(2i)^{n}} \lim_{\epsilon \to 0^{+}} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} 2 \pi i \ \frac{1}{(n-1)!} i^{n-1} (n-2k)^{n-1} e^{-\epsilon(n-k)}$

$ \displaystyle = \frac{\pi}{2^{n}(n-1)!} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{k} (-1)^{k} (n-2k)^{n-1} $

So, for example, $\displaystyle \int_{0}^{\infty} \frac{\sin^{5} x}{x^{5}} \ dx = \frac{\pi}{2^{5}4!} \sum_{k=0}^{2} \binom{5}{k} (-1)^{k}(5-2k)^{4}= \frac{\pi}{768} \Big(5^4-5(3)^{4} +10(1)^4 \Big) = \frac{115 \pi}{384}$

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- Thread starter
- #10

- Jan 17, 2013

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[tex]F(a)=\int^{\infty}_0 \frac{\sin(x) e^{-ax}}{x}\, dx[/tex]

[tex]F'(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx[/tex]

[tex]F'(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx=\frac{-1}{a^2+1}[/tex]

[tex]F(a)=-\arctan(a)+C[/tex]

[tex]C=\lim_{a\to \infty }F(a) +\arctan(a)= \frac{\pi}{2}[/tex]

[tex]F(a)=-\arctan(a)+ \frac{\pi}{2}[/tex]

[tex]F(0)=\int^{\infty}_0 \frac{\sin(x) }{x}=\frac{\pi}{2}[/tex]

- Jan 31, 2012

- 253

$\int_{0}^{\infty} \sin (x) e^{-ax} = \frac{1}{1+a^{2}} $ for $a >0$

So really what you're saying is that

$ \lim_{a \to 0^{+}} \int_{0}^{\infty} \frac{\sin (x) e^{-ax}}{x} \ dx = \int_{0}^{\infty} \lim_{a \to 0} \frac{\sin (x) e^{-ax}}{x} \ dx = \int_{0}^{\infty} \frac{\sin x}{x} \ dx = \lim_{a \to 0} \Big( \arctan(a) + \frac{\pi}{2} \Big) = \frac{\pi}{2}$

Bringing the limit inside of the integral is not easy to justify since $\int_{0}^{\infty} \frac{\sin x}{x} \ dx$ does not converge absolutely.

The following is an example where things go horribly wrong.

$\lim_{a \to 0^{+}} \int_{0}^{\infty} \sin(x) e^{-ax} \ dx = \int_{0}^{\infty} \lim_{a \to 0^{+}} \sin(x) e^{-ax} \ dx =\int_{0}^{\infty} \sin x \ dx = \lim_{a \to 0^{+}} \frac{1}{1+a^{2}}= 1 $

So really what you're saying is that

$ \lim_{a \to 0^{+}} \int_{0}^{\infty} \frac{\sin (x) e^{-ax}}{x} \ dx = \int_{0}^{\infty} \lim_{a \to 0} \frac{\sin (x) e^{-ax}}{x} \ dx = \int_{0}^{\infty} \frac{\sin x}{x} \ dx = \lim_{a \to 0} \Big( \arctan(a) + \frac{\pi}{2} \Big) = \frac{\pi}{2}$

Bringing the limit inside of the integral is not easy to justify since $\int_{0}^{\infty} \frac{\sin x}{x} \ dx$ does not converge absolutely.

The following is an example where things go horribly wrong.

$\lim_{a \to 0^{+}} \int_{0}^{\infty} \sin(x) e^{-ax} \ dx = \int_{0}^{\infty} \lim_{a \to 0^{+}} \sin(x) e^{-ax} \ dx =\int_{0}^{\infty} \sin x \ dx = \lim_{a \to 0^{+}} \frac{1}{1+a^{2}}= 1 $

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- Thread starter
- #12

- Jan 17, 2013

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If We take \(\displaystyle F(a)=\lim_{a \to 0^{+}} \int_{0}^{\infty} \frac{\sin (x) e^{-ax}}{x} \ dx \)$$ \lim_{a \to 0^{+}} \int_{0}^{\infty} \frac{\sin (x) e^{-ax}}{x} \ dx = \int_{0}^{\infty} \lim_{a \to 0} \frac{\sin (x) e^{-ax}}{x} \ dx = \int_{0}^{\infty} \frac{\sin x}{x} \ dx = \lim_{a \to 0} \Big( \arctan(a) + \frac{\pi}{2} \Big) = \frac{\pi}{2}$$

since F(a) exists for a=0 then we can pass the limit inside .

Since at a=0 the integral doesn't converge , we cannot pass the limit inside .The following is an example where things go horribly wrong.

$$\lim_{a \to 0^{+}} \int_{0}^{\infty} \sin(x) e^{-ax} \ dx = \int_{0}^{\infty} \lim_{a \to 0^{+}} \sin(x) e^{-ax} \ dx =\int_{0}^{\infty} \sin x \ dx = \lim_{a \to 0^{+}} \frac{1}{1+a^{2}}= 1 $$

- Jan 31, 2012

- 253

There is no theorem that says that's true in general.since F(a) exists for a=0 then we can pass the limit inside

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- #14

- Jan 17, 2013

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It always works for me , may be I need to prove it. Can you give an example that if we swap we get a different convergent value ?There is no theorem that says that's true in general.

- Jan 31, 2012

- 253

But $ \displaystyle \lim_{n \to \infty} \int_{0}^{\infty} \frac{n \arctan x}{n^2+x^{2}} \ dx$ tends to $\frac{\pi^{2}}{4}$

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- #16

- Mar 5, 2012

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Isn't this the same as

[tex]F(a)=\int^{\infty}_0 \frac{\sin(x) e^{-ax}}{x}\, dx[/tex]

[tex]F'(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx[/tex]

[tex]F'(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx=\frac{-1}{a^2+1}[/tex]

[tex]F(a)=-\arctan(a)+C[/tex]

[tex]C=\lim_{a\to \infty }F(a) +\arctan(a)= \frac{\pi}{2}[/tex]

[tex]F(a)=-\arctan(a)+ \frac{\pi}{2}[/tex]

[tex]F(0)=\int^{\infty}_0 \frac{\sin(x) }{x}=\frac{\pi}{2}[/tex]

Note that \(\displaystyle \mathcal L\{\frac {\sin x} x\}(a) = \int^{\infty}_0 \frac{\sin(x) e^{-ax}}{x}\, dx\).

Combine with the Laplace formula for taking a derivative and there you go...

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- #17

- Jan 17, 2013

- 1,667

Lots of things look different while they are actually the same. Finding the connection is not so easy, though .Isn't this the same aschisigma's solution?

Note that \(\displaystyle \mathcal L\{\frac {\sin x} x\}(a) = \int^{\infty}_0 \frac{\sin(x) e^{-ax}}{x}\, dx\).

Combine with the Laplace formula for taking a derivative and there you go...

- Jan 31, 2012

- 253

My preferred method

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x} \ dx$

Integrate by parts by letting $u= \frac{1}{x}$ and $dv = \sin x \ dx$.

For $v$ choose the antiderivative $1- \cos x$.

$ \displaystyle = \frac{1- \cos x}{x} \big|^{\infty}_{0} + \int_{0}^{\infty} \frac{1- \cos x}{x^{2}} \ dx = \int_{0}^{\infty} \frac{1- \cos x}{x^{2}} \ dx$

$ \displaystyle = \int_{0}^{\infty} \int_{0}^{\infty} (1- \cos x) te^{-xt} \ dt \ dx $

Since the integrand is always nonnegative, Fubini's/Tonelli's theorem says that we can switch the order of integration. That's why I initially integrated by parts.

$ \displaystyle = \int_{0}^{\infty} \int_{0}^{\infty} t (1- \cos x) e^{-tx} \ dx \ dt = \int_{0}^{\infty} t \Big( \frac{1}{t} - \frac{t}{1+t^{2}} \Big) \ dt$

$ \displaystyle = \int_{0}^{\infty} \frac{1}{1+t^{2}} \ dt = \frac{\pi}{2}$

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x} \ dx$

Integrate by parts by letting $u= \frac{1}{x}$ and $dv = \sin x \ dx$.

For $v$ choose the antiderivative $1- \cos x$.

$ \displaystyle = \frac{1- \cos x}{x} \big|^{\infty}_{0} + \int_{0}^{\infty} \frac{1- \cos x}{x^{2}} \ dx = \int_{0}^{\infty} \frac{1- \cos x}{x^{2}} \ dx$

$ \displaystyle = \int_{0}^{\infty} \int_{0}^{\infty} (1- \cos x) te^{-xt} \ dt \ dx $

Since the integrand is always nonnegative, Fubini's/Tonelli's theorem says that we can switch the order of integration. That's why I initially integrated by parts.

$ \displaystyle = \int_{0}^{\infty} \int_{0}^{\infty} t (1- \cos x) e^{-tx} \ dx \ dt = \int_{0}^{\infty} t \Big( \frac{1}{t} - \frac{t}{1+t^{2}} \Big) \ dt$

$ \displaystyle = \int_{0}^{\infty} \frac{1}{1+t^{2}} \ dt = \frac{\pi}{2}$

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- #19

- Mar 5, 2012

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I guess the trick is in understanding how the Laplace transform works.Lots of things look different while they are actually the same. Finding the connection is not so easy, though .

Once you learn how to apply it in its integral form, your derivation pops out.

As for the Fourier transform, I have to admit that it's not so nice to look up a transform.

On the other hand, it's really easy to deduce the inverse transform of the rectangle function:

$$\mathcal F^{-1}\{\text{rect } \xi\}(x) = \int_{-\infty}^\infty \text{rect } \xi \ e^{2\pi i \xi x} d\xi = \int_{-1/2}^{1/2} e^{2\pi i \xi x} d\xi = \left. \frac 1 {2\pi i x} e^{2\pi i \xi x} \right|_{-1/2}^{1/2} = \frac 1 {2i \pi x}(e^{i \pi x} - e^{-i \pi x}) = \text{sinc x}$$

So that makes the Fourier transform solution still pretty easy and really different (that is, no derivative involved).

- - - Updated - - -

Where did that new integral come from?$ \displaystyle = \int_{0}^{\infty} \int_{0}^{\infty} (1- \cos x) te^{-xt} \ dt \ dx $

It looks suspiciously like a Laplace transform.

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- Jan 31, 2012

- 253

I just expressed the integral as an iterated integral by using the Laplace transform $ \displaystyle \int_{0}^{\infty} t e^{-xt} \ dt = \frac{1}{x^{2}}$.Where did that new integral come from?

It looks suspiciously like a Laplace transform.

In general, $\displaystyle \int_{0}^{\infty} t^{n} e^{-xt} \ dt = \frac{n!}{x^{n+1}}$.

A lot of integrals can be evaluated in this manner.

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