Prove one of the following - Part Deux

[DreamWeaver]

Problem 3:

$$\int_0^{\infty} \frac{\sin^{-1}(b\sin x)}{x}\, dx=\text{Si}_2(b)$$
Where$$\text{Si}_1(x)=\sin^{-1}x$$$$\text{Si}_{m+1}(x)=\int_0^x\frac{\text{Si}_m(t)}{t}\,dt$$
Problem 4:$$\int_0^{\infty}\frac{\text{Si}_m(b\sin x)}{x}\,dx=\text{Si}_{m+1}(b)$$

 

[jvicens]

Hello,

Thank you for sharing this interesting problem. The solution to Problem 3 can be found by using the definition of the Si function and integrating by parts:

\begin{align*}
\int_0^{\infty} \frac{\sin^{-1}(b\sin x)}{x}\, dx &= \int_0^{\infty} \frac{\text{Si}_1(b\sin x)}{x}\, dx\\
&= \int_0^{\infty} \frac{1}{x} \int_0^{b\sin x} \frac{\sin^{-1}t}{t}\, dt\, dx\\
&= \int_0^{\infty} \frac{1}{t} \int_0^x \frac{\sin^{-1}t}{t}\, dx\, dt\\
&= \int_0^{\infty} \frac{\sin^{-1}t}{t}\, dt\\
&= \text{Si}_2(b)
\end{align*}

For Problem 4, we can use induction to prove the result. For m = 1, we have already shown the result in Problem 3. Now, assuming that the result holds for m, we can use integration by parts again to show that it also holds for m+1:

\begin{align*}
\int_0^{\infty} \frac{\text{Si}_{m+1}(b\sin x)}{x}\, dx &= \int_0^{\infty} \frac{1}{x} \int_0^{b\sin x} \frac{\text{Si}_m(t)}{t}\, dt\, dx\\
&= \int_0^{\infty} \frac{1}{t} \int_0^x \frac{\text{Si}_m(t)}{t}\, dx\, dt\\
&= \int_0^{\infty} \frac{\text{Si}_m(t)}{t}\, dt\\
&= \text{Si}_{m+1}(b)
\end{align*}

Therefore, the result holds for all natural numbers m. I hope this helps with understanding these integrals. Let me know if you have any further questions.