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Integral

Posted December 4th, 2007 by Krizalid
in

Evaluate

$$\int_0^\infty {\frac{{2 - 2\cos x}}<br /> {{xe^x }}\,dx} .$$

It's a nice problem, give it a try!

‹ second derivative

Comments

Okay, here's another

On December 7th, 2007 Krizalid says:

Okay, here's another solution:

$ </p> <p>\begin{eqnarray*}<br /> \int_0^\infty {\frac{{2 - 2\cos x}}<br /> {{xe^x }}\,dx} &=& 2\int_0^\infty {\int_0^1 {e^{ - x} \sin (ux)\,du} \,dx}\\<br /> &=& 2\int_0^1 {\int_0^\infty {e^{ - x} \sin (ux)\,dx} \,du}\\<br /> &=& 2\int_0^1 {\operatorname{Im} \left[ {\int_0^\infty {e^{ - (1 - iu)x} \,dx} } \right]\,du}\\<br /> &=& 2\int_0^1 {\operatorname{Im} \frac{1}<br /> {{1 - iu}}\,du} \hfill \\<br /> &=& \int_0^1 {\frac{{2u}}<br /> {{1 + u^2 }}\,du}\\<br /> &=& \ln 2\,\blacksquare<br /> \end{eqnarray*} $

On Fubini theorem

On December 7th, 2007 Structure says:

Nice work.
Thanks for sharing your solution. Fubini is essential here and in fact both our solutions use an integral equivalent to the Laplace transform of the sin function.

Laplace method

On December 5th, 2007 Structure says:

Integral can be written

$$2\int_{0}^{+\infty}\frac{1-\cos x}{x}e^{-x}dx$$

This form suggest Laplace transform.

$$\mathcal{L}(f(x))(s)=\int_{0}^{+\infty}f(x)e^{-sx}dx$$

We have

$$\mathcal{L}(1-\cos x)(s)=\int_{0}^{+\infty}(1-\cos x)e^{-sx}dx=\frac{1}{s}-\frac{s}{s^2+1}$$

We have from here

$$\mathcal{L}(\frac{1-\cos x}{x})(s)=\int_{0}^{+\infty}(\frac{1-\cos x}{x})e^{-sx}dx=\ln \frac{\sqrt {s^2+1}}{s}$$

So

$$2\int_{0}^{+\infty}\frac{1-\cos x}{x}e^{-x}dx=2\mathcal{L}(\frac{1-\cos x}{x})(1)=\ln 2$$

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