Let $a \geq 0$ . Prove that the integral $\displaystyle \int_0^\infty\frac{\sin (x^2+ax)}{x}\, {\rm d}x$.
Solution
We first note that $\displaystyle \lim_{x\rightarrow 0}f(x)= \lim_{x\rightarrow 0}\frac{\sin (x^2+ax)}{x}=a$ meaning that $f$ is locally Riemann integrable in $[0, +\infty)$, that is in every interval $\Delta \subseteq [0, +\infty)$. In order for the integral to exist we must have according to Cauchy's criterion that:
$$\lim_{x_1 \rightarrow +\infty} \lim_{x_2 \rightarrow +\infty}\left | \int_{x_1}^{x_2}\frac{\sin \left ( x^2+ax \right )}{x} \right |\, {\rm d}x =0$$
However,
$$\begin{aligned}
\left| {\int\limits_{{x_1}}^{{x_2}} {\frac{{\sin \left( {{x^2} + ax} \right)}}{x}\, {\rm d}x} } \right| &= \left| {\int\limits_{{x_1}}^{{x_2}} {\frac{{\cos \left( {{x^2} + ax} \right)'}}{{\left( {2x + a} \right)x}}\;{\rm d}x} } \right|\\
&= \left| {\left[ {\frac{{\cos \left( {{x^2} + ax} \right)}}{{\left( {2x + a} \right)x}}} \right]_{{x_1}}^{{x_2}} + \int\limits_{{x_1}}^{{x_2}} {\frac{{4ax + {a^2}}}{{{x^2}{{\left( {2x + a} \right)}^2}}}\cos \left( {{x^2} + ax} \right)\;{\rm d}x} } \right| \\
&= \left| {\frac{{\cos \left( {x_2^2 + a{x_2}} \right)}}{{\left( {2{x_2} + a} \right){x_2}}} - \frac{{\cos \left( {x_1^2 + a{x_2}} \right)}}{{\left( {2{x_1} + a} \right){x_1}}} + \\ \quad \quad \quad \quad \int\limits_{{x_1}}^{{x_2}} {\frac{{\left( {4ax + {a^2}} \right)\cos \left( {{x^2} + ax} \right)}}{{{x^2}{{\left( {2x + a} \right)}^2}}}\, {\rm d}x} } \right| \\
&\leq \frac{{\left| {\cos \left( {x_2^2 + a{x_2}} \right)} \right|}}{{\left( {2{x_2} + a} \right){x_2}}} + \frac{{\left| {\cos \left( {x_1^2 + a{x_2}} \right)} \right|}}{{\left( {2{x_1} + a} \right){x_1}}} +\\
&\quad \quad \quad \quad \quad + \left| {\int\limits_{{x_1}}^{{x_2}} {\frac{{\left( {4ax + {a^2}} \right)\cos \left( {{x^2} + ax} \right)}}{{{x^2}{{\left( {2x + a} \right)}^2}}}\, {\rm d}x} } \right| \\
& \leq \frac{1}{{\left( {2{x_2} + a} \right){x_2}}} + \frac{1}{{\left( {2{x_1} + a} \right){x_1}}} + \\
& \quad \quad \quad \quad \quad \quad \quad +\int\limits_{{x_1}}^{{x_2}} {\frac{{\left( {4ax + {a^2}} \right)\left| {\cos \left( {{x^2} + ax} \right)} \right|}}{{{x^2}{{\left( {2x + a} \right)}^2}}}\, {\rm d}x} \\
&= \frac{1}{{\left( {2{x_2} + a} \right){x_2}}} + \frac{1}{{\left( {2{x_1} + a} \right){x_1}}} + \int\limits_{{x_1}}^{{x_2}} {\frac{1}{{{x^2}}}\, {\rm d}x} \\
&= \frac{1}{{\left( {2{x_2} + a} \right){x_2}}} + \frac{1}{{\left( {2{x_1} + a} \right){x_1}}} + \frac{1}{{{x_1}}} - \frac{1}{{{x_2}}} \xrightarrow{x_1 \rightarrow +\infty, \\ x_2 \rightarrow +\infty}0
\end{aligned}$$
meaning that the integral converges.
(Qualifying Exams, Wisconsin / Madison University, 2015)
Solution
We first note that $\displaystyle \lim_{x\rightarrow 0}f(x)= \lim_{x\rightarrow 0}\frac{\sin (x^2+ax)}{x}=a$ meaning that $f$ is locally Riemann integrable in $[0, +\infty)$, that is in every interval $\Delta \subseteq [0, +\infty)$. In order for the integral to exist we must have according to Cauchy's criterion that:
$$\lim_{x_1 \rightarrow +\infty} \lim_{x_2 \rightarrow +\infty}\left | \int_{x_1}^{x_2}\frac{\sin \left ( x^2+ax \right )}{x} \right |\, {\rm d}x =0$$
However,
$$\begin{aligned}
\left| {\int\limits_{{x_1}}^{{x_2}} {\frac{{\sin \left( {{x^2} + ax} \right)}}{x}\, {\rm d}x} } \right| &= \left| {\int\limits_{{x_1}}^{{x_2}} {\frac{{\cos \left( {{x^2} + ax} \right)'}}{{\left( {2x + a} \right)x}}\;{\rm d}x} } \right|\\
&= \left| {\left[ {\frac{{\cos \left( {{x^2} + ax} \right)}}{{\left( {2x + a} \right)x}}} \right]_{{x_1}}^{{x_2}} + \int\limits_{{x_1}}^{{x_2}} {\frac{{4ax + {a^2}}}{{{x^2}{{\left( {2x + a} \right)}^2}}}\cos \left( {{x^2} + ax} \right)\;{\rm d}x} } \right| \\
&= \left| {\frac{{\cos \left( {x_2^2 + a{x_2}} \right)}}{{\left( {2{x_2} + a} \right){x_2}}} - \frac{{\cos \left( {x_1^2 + a{x_2}} \right)}}{{\left( {2{x_1} + a} \right){x_1}}} + \\ \quad \quad \quad \quad \int\limits_{{x_1}}^{{x_2}} {\frac{{\left( {4ax + {a^2}} \right)\cos \left( {{x^2} + ax} \right)}}{{{x^2}{{\left( {2x + a} \right)}^2}}}\, {\rm d}x} } \right| \\
&\leq \frac{{\left| {\cos \left( {x_2^2 + a{x_2}} \right)} \right|}}{{\left( {2{x_2} + a} \right){x_2}}} + \frac{{\left| {\cos \left( {x_1^2 + a{x_2}} \right)} \right|}}{{\left( {2{x_1} + a} \right){x_1}}} +\\
&\quad \quad \quad \quad \quad + \left| {\int\limits_{{x_1}}^{{x_2}} {\frac{{\left( {4ax + {a^2}} \right)\cos \left( {{x^2} + ax} \right)}}{{{x^2}{{\left( {2x + a} \right)}^2}}}\, {\rm d}x} } \right| \\
& \leq \frac{1}{{\left( {2{x_2} + a} \right){x_2}}} + \frac{1}{{\left( {2{x_1} + a} \right){x_1}}} + \\
& \quad \quad \quad \quad \quad \quad \quad +\int\limits_{{x_1}}^{{x_2}} {\frac{{\left( {4ax + {a^2}} \right)\left| {\cos \left( {{x^2} + ax} \right)} \right|}}{{{x^2}{{\left( {2x + a} \right)}^2}}}\, {\rm d}x} \\
&= \frac{1}{{\left( {2{x_2} + a} \right){x_2}}} + \frac{1}{{\left( {2{x_1} + a} \right){x_1}}} + \int\limits_{{x_1}}^{{x_2}} {\frac{1}{{{x^2}}}\, {\rm d}x} \\
&= \frac{1}{{\left( {2{x_2} + a} \right){x_2}}} + \frac{1}{{\left( {2{x_1} + a} \right){x_1}}} + \frac{1}{{{x_1}}} - \frac{1}{{{x_2}}} \xrightarrow{x_1 \rightarrow +\infty, \\ x_2 \rightarrow +\infty}0
\end{aligned}$$
meaning that the integral converges.
As a matter of fact , not only does the integral converges but it also defines a continuous function with respect to $a$. That is, the function:
ReplyDelete$$f(a)=\int_0^\infty \frac{\sin (x^2+ax)}{x}\, {\rm d}x$$
is continuous. Indeed , using the Liebniz differentiation we have that:
$$\begin{aligned}
f'(a) &=\int_{0}^{\infty}\cos \left ( x^2+ax \right )\, {\rm d}x\int_{0}^{\infty}\cos \left [ \left ( x+\frac{a}{2} \right )^2 - \frac{a^2}{4} \right ]\, {\rm d}x \\
&= \cos \frac{a^2}{4}\int_{0}^{\infty}\cos \left ( x+ \frac{a}{2} \right )^2\, {\rm d}x + \sin \frac{a^2}{4}\int_{0}^{\infty}\sin \left ( x+ \frac{a}{2} \right )\, {\rm d}x \\
&= \cos \frac{a^2}{4}\int_{a/2}^{\infty} \cos x^2 \, {\rm d}x + \sin \frac{a^2}{4}\int_{a/2}^{\infty}\sin x^2 \, {\rm d}x\\
&=\cos \frac{a^2}{4} \int_{0}^{\infty} \cos x^2 \, {\rm d}x +\sin \frac{a^2}{4}\int_{0}^{\infty}\sin x^2 \, {\rm d}x - \\
& \quad \qquad \qquad \qquad - \left ( \cos \frac{a^2}{4}\int_{0}^{a/2}\cos x^2 \, {\rm d}x + \sin \frac{a^2}{4}\int_{0}^{a/2}\sin x^2\, {\rm d}x \right )
\end{aligned}$$
Quoting the Fresnel integrals we see that $f$ is differentiable, hence continuous as we suggested in the very beginning.
That was the second part of the question of the original exams.