Evaluating $ \int_0^4 \frac{\sin \sqrt{t}}{\sqrt{t}} dt$

Question

Evaluate the integral given below: $$ \int_0^4 \frac{\sin \sqrt{t}}{\sqrt{t}} dt$$

I tried evaluating this integral by using substitution theorem for integrals( by defining $f(x)= 2\sin(x)$ and $\Phi(t)= t^{1/2}$ but turns out when defining $\Phi$, the derivative doesn't exist at zero and so I can't really apply the substitution here.

And so I thought if I could apply Fundamental theorem for calculus on the function $f(x)$, that $f$ is integrable for $(0,2]$ where $0=\Phi(0)$ and $2=\Phi(4)$, but then I got lost. I remember reading somewhere that if the function is bounded by $M$ and integrable on restricted intervals $[c,b]$ where $c$ is an interior point in $(a,b)$ then it becomes integrable on the whole interval $[a,b]$, that's what I thought will help me to apply Fundamental theorem (I think squeeze theorem might come in handy as well) but I still have no idea how will it help me to evaluate the value of the integral? I seem to be missing a lot here. Can I get a helpful hint or a solution to this problem?

P.S: I know the solution where I could simply use precalculus technique of substitution to solve but I need to prove that the integral exists using fundamental theorem for calculus given in real analysis.

Answer 1

When integrating over $(a,b)$, you are really concerned with what happens inside the integral. In other words, write $$ \int_a^b f(x)dx = \lim_{A \to a+} \int_A^b f(x) dx, $$ where $A > a$ and use your substitution $\Phi(t) = \sqrt{t}$. Even though $\Phi'(a)$ does not exist, still, $\Phi'(t)$ does exist anywhere on $[A,b]$ with $A>a$ and you are integrating over $[A,b]$ first and taking the limit as $A \to a$ only afterwards.

Answer 2

The derivative of the substutiting function does not need to exist at the domain endpoints – all that matters is for it to be bijective over the domain. Clearly $\sqrt t$ is bijective over $[0,4]$, so the integral evaluates to $$2\int_0^2\sin u\,du=2(1-\cos2)$$

Answer 3

To show integrability, just note

$$\lim_{t\to 0^+} \frac{\sin \sqrt t}{\sqrt t} = 1,$$

and thus, after defining the integrand to be $1$ at $0,$ you have a continuous function on $[0,4].$ Then you can go ahead and let $t=s^2$  and everything works out just fine.