Differentiation under the integral sign, where the partial derivative of the integrand is not bounded by a Lebesgue integrable function.

Question

Let $K(t)=\int_1^\infty u(t,x)\ \mathrm{d}x$, where $$u(t,x)=\frac{\cos{tx}}{x^2}\mathbb{1}_{[1,\infty)}(x).$$ I need to show that, for $t>0$, $$\frac{dK}{dt}(t)=\frac{1}{t}\left(K(t)-\cos{t}\right).$$ I have shown that $\int_1^\infty \frac{\partial}{\partial t}u(t,x)\ \mathrm{d}x$ is equal to the right-hand side of the above, but I cannot use the standard theorem for differentiation under the integral sign to establish that $$\int_1^\infty \frac{\partial}{\partial t}u(t,x)\ \mathrm{d}x=\frac{dK}{dt}(t),$$ since $$\frac{\partial}{\partial t}u(t,x)=\frac{-\sin{tx}}{x}$$ is not bounded above by a Lebesgue integral function. How do I therefore show that the equality still works?

Answer 1

Let us start instead by integrating by parts. Assume that $t>0$. We find that $$ \begin{aligned} K(t)&=\int_1^{+\infty}\frac{1}{x^2}\cos tx\,dx=\Bigl[\frac{\sin tx}{t}\frac{1}{x^2}\Bigr]_1^{+\infty}+\int_1^{+\infty}\frac{\sin tx}{t}\frac{2}{x^3}\,dx\\ &=-\frac{\sin t}{t}+\frac{2}{t}\int_1^{+\infty}\frac{\sin tx}{x^3}\,dx \end{aligned} $$ Now, we can differentiate (with your standard theorem), and we find that $$ \begin{aligned} K'(t)&=-\frac{t\cos t-\sin t}{t^2}-\frac{2}{t^2}\int_1^{+\infty}\frac{\sin tx}{x^3}+\frac{2}{t}\int_1^{+\infty}\frac{\cos tx}{x^2}\,dx\\ &=\frac{1}{t}\bigl(-\cos t-K(t)+2K(t)\bigr)\\ &=\frac{1}{t}\bigl(K(t)-\cos t\bigr), \end{aligned} $$ and then we are done.