This problem is about the integral
\[I = \int_0^\infty \frac{\cos u \, du}{u^4 + 1}.\]
Show directly that \(I\) is a convergent improper Riemann integral.
Is
\[\int_{[0,\infty)} \frac{\cos u}{u^4 + 1} \, d\mu(u)\]
a well-defined Lebesgue integral, where \(\mu\) denotes the Lebesgue measure on \(\mathbb{R}\)?
- (main part) Evaluate the integral in (a).
Solution
Both integrals exist, agree, and have the value \[ \boxed{\frac{\pi e^{-1/\sqrt2}}{2\sqrt2} \left(\cos\frac1{\sqrt2}+\sin\frac1{\sqrt2}\right)}. \]
1. Parts (a) and (b): absolute convergence holds in both senses.
The function \(h(u)=\cos u/(1+u^4)\) is continuous on \([0,\infty)\). It is bounded by one in absolute value on \([0,1]\) and by \(u^{-4}\) for \(u\geq1\). Since \(\int_1^\infty u^{-4}\,du=1/3\), comparison proves absolute convergence of the improper Riemann integral. The same bounds prove \(h\in L^1([0,\infty),\mu)\), since continuity implies measurability [@Fol13]. On each finite interval the Riemann and Lebesgue integrals coincide; dominated convergence, with the displayed integrable majorant, shows that their infinite-interval limits agree.
2. The exponential integrand has two relevant simple poles.
Consider \(F(z)=e^{iz}/(1+z^4)\). The zeros of the denominator in the upper half-plane are \[ \alpha=\frac{1+i}{\sqrt2},\qquad \beta=\frac{-1+i}{\sqrt2}. \] They are simple because \(4z^3\) is nonzero there, so the residue sum is \(S=e^{i\alpha}/(4\alpha^3)+e^{i\beta}/(4\beta^3)\) [@SS03]. Set \(s=1/\sqrt2\). Then \[ \alpha^{-3}=\frac{-1-i}{\sqrt2},\quad \beta^{-3}=\frac{1-i}{\sqrt2},\quad e^{i\alpha}=e^{-s}e^{is},\quad e^{i\beta}=e^{-s}e^{-is}. \] It follows that \[ S=\frac{e^{-s}}{4\sqrt2} \bigl((-1-i)e^{is}+(1-i)e^{-is}\bigr) =-\frac{ie^{-s}}{2\sqrt2}(\cos s+\sin s). \]
3. Part (c): the upper semicircle evaluates the real integral.
1. Letting \(R\to\infty\)and taking real parts therefore gives\[ \int_{-\infty}^\infty\frac{\cos u}{1+u^4}\,du =\frac{\pi e^{-s}}{\sqrt2}(\cos s+\sin s). \]Evenness makes the left side equal to \(2I\). Dividingby two and substituting \(s=1/\sqrt2\) proves the answer.