f(k) - f(1)
= $\int_{-\infty }^{k}\frac{e^{sinx}}{x}$ - $\int_{-\infty}^{1}\frac{e^{sinx}}{x}$
= $\int_{1}^{k}\frac{e^{sinx}}{x}$ ...................(1)
Now, we simplify the integral
I = $\int_{1}^{4}\frac{2e^{sinx^{2}}}{x}$
Put x^{2} = t
$\therefore 2x dx=dt$
$\therefore dx=\frac{dt}{2\sqrt{t}}$
$\therefore$ I = $\int_{1}^{16} \frac{2e^{sint}dt}{\sqrt{t}*2\sqrt{t}}$ ...........(2)
Now, when we simplify statement (2), we get statement (1).
$\therefore k=16$