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Prove the following equation:

$$\int_0^{\infty} \frac{\cos{(x)}}{1+x} \,\mathrm dx=\int_0^{\infty} \frac{\sin{(x)}}{(1+x)^2} \,\mathrm dx$$

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By integration by parts we have $$\int_0^A\frac{\cos x}{1+x}dx=\left[\frac{\sin x}{1+x}\right]_0^A+\int_0^A\frac{\sin x}{(1+x)^2}dx$$ and with the inequality $|\sin x|\leq 1$ and passing to limit $A\to\infty$ we find the desired result.

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Mark: $1+x=t \to x=t-1$ $$\int_1^{\infty} \frac{\cos{(t-1)}}{t} \,\mathrm dt$$

Now integration by parts: $u=\frac1t, u'=-\frac{1}{t^2}, v'=\cos(t-1), v=\sin(t-1)$

And i get: $-\frac1t\sin(t-1)|_1^\infty+\int_1^{\infty} \frac{\sin{(t-1)}}{t^2} \,\mathrm dt$

Since: $-\frac1t\sin(t-1)|_1^\infty$ equals $o$, Place $t=1+x$ in the equation again, and we get: $$\int_0^{\infty} \frac{\sin{(x)}}{(1+x)^2} \,\mathrm dx$$

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