Differentiate under integral sign

Going by the title and the use of 1729 as the co-efficient of x, people might be tempted to assign a random variable for 1729 and differentiate with respect to that (I was too.). But that won't help us as we end up getting indeterminate values to deal with in the end. The trick is to cancel out the 'x' in the denominator and to do it in such a way that in the end the bounds given won't make us regret approaching it that way. This is my approach: Let us call 1729 as 'a' and let b be any number. $I\left( b \right) =\int _{ 0 }^{ \infty }{ \frac { sin(ax) }{ x } { e }^{ -bx } } dx\\ Differentiate\quad with\quad respect\quad to\quad b\\ I^{ ' }\left( b \right) =\int _{ 0 }^{ \infty }{ -sin(ax){ e }^{ -bx } } dx\\ sin(ax)\quad can\quad be\quad written\quad as\quad Im\{ { e }^{ iax }\} \quad (Imaginary\quad part\quad of\quad { e }^{ iax })\\ I^{ ' }\left( b \right) =Im\{ \int _{ 0 }^{ \infty }{ -{ e }^{ -(b-ia)x } } dx\} \\ I^{ ' }\left( b \right) =Im\{ \frac { 1 }{ b-ia } [{ e }^{ (b-ia)(-\infty ) }-{ e }^{ (b-ia)(0) }]\} \\ I^{ ' }\left( b \right) =Im\{ \frac { -1 }{ b-ia } \} \\ Multiplying\quad and\quad dividing\quad by\quad (b+ia)\quad we\quad get:\\ I^{ ' }\left( b \right) =Im\{ \frac { -b-ia }{ { b }^{ 2 }+{ a }^{ 2 } } \} \\ Taking\quad only\quad the\quad imaginary\quad part,\quad we\quad get\\ I^{ ' }\left( b \right) =\frac { -a }{ { b }^{ 2 }+{ a }^{ 2 } } \\ Integrating\quad both\quad sides\quad with\quad respect\quad to\quad b\\ I\left( b \right) =\int { \frac { -a }{ { b }^{ 2 }+{ a }^{ 2 } } db } \\ I\left( b \right) =-{ tan }^{ -1 }(\cfrac { b }{ a } )\quad +c\\ Since\quad I(\infty )=0\\ c=\frac { \pi }{ 2 } \\ Inserting\quad b=0\quad and\quad a=1729\quad back\quad into\quad the\quad equation:\\ I(0)=-{ tan }^{ -1 }(\cfrac { 0 }{ 1729 } )+\frac { \pi }{ 2 } =\boxed{\frac { \pi }{ 2 }}$ Note: I'm still familiarizing myself with this new method of differentiating under the integral. If I've gone wrong somewhere, please do point it out.

This all easy, $I=\int_{0}^{\infty} \dfrac{\sin(1729x)}{x} \, dx$

Now, let $y=1729x$ . So, $dx = \dfrac{dy}{1729}$ and $\dfrac1x = \dfrac{1729}y$

So $\begin{aligned}I&= \int_{0}^{\infty} \dfrac{\sin(1729x)}{x} \, dx\\&=\int_{0}^{\infty} \dfrac{1729\sin y}{y} \, \dfrac{dy}{1729}\\&=\int_{0}^{\infty} \dfrac{\sin y}{y} \, dy\end{aligned}$

Now use Dirichlet Integral $\int_{0}^{\infty} \dfrac{\sin\omega}{\omega} \, d\omega = \dfrac\pi2$

So $I = \int_{0}^{\infty} \dfrac{\sin(1729x)}{x} \, dx = \dfrac\pi2$

Proof for Dirichlet Integral, $\begin{aligned}I(a,x) &=\int_{0}^{\infty}e^{-ax}\dfrac{\sin x}xdx\\\dfrac{dI}{da}&= \int_{0}^{\infty}\dfrac\partial{\partial a}e^{-ax}\dfrac{\sin x}xdx\\&=-\Im \int_{0}^{\infty}\dfrac\partial{\partial a}e^{-ax}e^{ix}dx\\&=\Im\dfrac1{-a+i}\\&=\Im\dfrac{-a-i}{a^2+1}\\&=\dfrac{-1}{a^2+1}\\\Rightarrow I &= \int \dfrac{-1}{a^2+1}\\&=-\tan^{-1}a+c\\\\\text{Now, }I(\infty,x)&= 0\\\Rightarrow 0&=-\tan^{-1}(\infty) + c \\\Rightarrow c &= \dfrac\pi2 \\\\\therefore \int_{0}^{\infty} \dfrac{\sin\omega}{\omega} \, d\omega &= I(0,x)= \Large\boxed{\dfrac\pi2}\end{aligned}$

Dirichlet Integral method is nice