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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

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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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