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I would like to show that

$$\int_0^{\infty}\frac{\sin x}{x}e^{-nx} \, dx=\arctan\frac{1}{n}$$

Any help is appreciate!

Michael Hardy
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Barbara
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1 Answers1

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HINT: Let $$I(n)=\int_0^{\infty}\dfrac{\sin x}{x}e^{-nx}\mathrm{d}x$$

then $$I'(n)=\int_0^{\infty}\dfrac{\sin x}{x}\left(\frac{\partial e^{-nx}}{\partial n}\right)\mathrm{d}x=-\int_0^{\infty}\sin x e^{-nx}\mathrm{d}x\tag{1}$$

EDIT:

Now, $$S=\int_0^{\infty}\sin x e^{-nx}\mathrm{d}x=\dfrac{\sin xe^{-nx}}{n}|_0^{\infty}+\frac{1}{n}\int_0^{\infty}\cos x e^{-nx}\mathrm{d}x\tag{using by parts}$$ $$=0+\frac{1}{n}\left(\frac{\cos x e^{-nx}}{n}|_0^{\infty}-\frac{1}{n}\int_0^{\infty}\sin x e^{-nx}\mathrm{d}x\right)$$ $$=\frac{1}{n}\left(\frac{1}{n}-\frac{S}{n}\right)=\frac{1}{n^2}-\frac{S}{n^2}$$ $$\implies S\left(1+\frac{1}{n^2}\right)=\frac{1}{n^2}$$ $$\implies S=\frac{1}{n^2+1}$$

Putting it in $(1)$ gives,

$$I'(n)=-S=-\frac{1}{n^2+1}$$ $$\implies I(n)=-\arctan(n)+c\tag{c is constant of inte.}$$

As $I(0)= \int_0^{\infty}\dfrac{\sin x}{x}\mathrm{d}x=\frac{\pi}{2}\implies I(n)=\frac{\pi}{2}-\arctan(n)=\arctan(\frac{1}{n})$

Aang
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  • Thank you very much! I have shown that $I(n)=-\arctan(n) +C$ and than that $\lim_{n\to\infty}I(n)=0$ so $C=\frac{\pi}{2}$. I have a question: Under what condition on $g(x,t)$ can one prove that $$\frac{d}{dt}\int_a^bg(x,t)dx=\int_a^b\frac{\partial g}{\partial t}g(x,t)dx$$ ? Can someone recommend me some links? Thank you. – Barbara Jul 14 '13 at 13:08