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

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.

• $u=\sqrt{t}$  Jun 7, 2021 at 14:26
• The derivative exists almost everywhere and the behavior at zero does not prevent the integral from converging.
– user65203
Jun 7, 2021 at 14:34
• @mulaan_20: as Yves Duoust and zhw. mentioned in their comments and answers, the existence of the integral is not an issue. You integrand is actually a continuous function (modulo removal of singularity). To evaluate it, one ay have to do it numerically, as the integrand does not have a primitive in the sense of Liouville Jun 7, 2021 at 20:45

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.
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 $$2\int_0^2\sin u\,du=2(1-\cos2)$$
$$\lim_{t\to 0^+} \frac{\sin \sqrt t}{\sqrt t} = 1,$$
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.