Integral

**liyi** · October 12th 2007, 07:45 PM

Evaluate $\int_0^\infty\frac{e^{-x}\sin x}x\,dx$

**Krizalid** · October 12th 2007, 07:58 PM

Fact you should know

$\int_0^\infty e^{-x}\cos(\alpha x)\,dx=\frac1{\alpha^2+1}$

Now, let's introduce the following parameter to construct a double integral as follows:

$\int_0^1\cos(\alpha x)\,d\alpha=\frac{\sin x}x,$ this yields

$\int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^\infty e^{-x}\left(\int_0^1\cos(\alpha x)\,d\alpha\right)\,dx$

Reversing the order of integration

$\int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^1\int_0^\infty e^{-x}\cos(\alpha x)\,dx\,d\alpha=\int_0^1\frac1{\alpha^2+1}\,d\alph a=\frac\pi4\,\blacksquare$

**topsquark** · October 13th 2007, 03:36 AM

Originally Posted by Krizalid

Fact you should know

$\int_0^\infty e^{-x}\cos(\alpha x)\,dx=\frac1{\alpha^2+1}$

Now, let's introduce the following parameter to construct a double integral as follows:

$\int_0^1\cos(\alpha x)\,d\alpha=\frac{\sin x}x,$ this yields

$\int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^\infty e^{-x}\left(\int_0^1\cos(\alpha x)\,d\alpha\right)\,dx$

Reversing the order of integration

$\int_0^\infty\frac{e^{-x}\sin x}x\,dx=\int_0^1\int_0^\infty e^{-x}\cos(\alpha x)\,dx\,d\alpha=\int_0^1\frac1{\alpha^2+1}\,d\alph a=\frac\pi4\,\blacksquare$

Impressive, as always!

-Dan

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