I am asked to show that
$\int_{0}^{\pi/2}(\sin^{m-1}x)(\cos^{n-1}x) dx\displaystyle \qquad$ is convergent for $m>0,n>0$
My understanding :
Let $(\sin^{m-1}x)(\cos^{n-1}x) = f(x)...(say)$
$\int_{0}^{\pi/2}f(x)dx = \underbrace{\int_{0}^{\pi/4}f(x)dx}_{\text{$I_{1}$}} + \underbrace{\int_{\pi/4}^{\pi/2}f(x)dx}_{\text{$I_{2}$}} $
For $I_{1}$:
So convergence at $x=0$ is contested due to $\sin x$ component which by comparison test with $\phi(x) = \frac{1}{x^{1-m}} $ shows that it converges for $1-m<1 => m > 0$
For $I_{2}$ :
Convergence at $x=\pi/2$ is contested due to $\cos x$ component. I tried to use definite integral formula $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$
which gives $I_{2} = \int_{0}^{\pi/2}(\cos^{m-1}x)(\sin^{n-1}x) dx$ which is of the form $I_{1}$ and it converges for $n>0$
My question :
- can we use definite integral formula here $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$ before we show that $\int_{a}^{b}f(x) dx$ is convergent ?
- If not or otherwise, suggest any other method
We can make the substitution $x=g(t)$ where $g$ is a strictly monotonic function with a continuous derivative in an improper integral: $$\int_a^bf(x)\,dx=\int_{\alpha}^{\beta}f(g(t))g'(t)\,dt $$ (The two integrals should have matching singularities at one of the integration limits). This holds provided one of the integrals is convergent, and its convergence implies the convergence of the other one. In other words, you can make a substitution, show that the obtained integral is convergent, and conclude that the original one is convergent as well. That answers your first question.
However, I don't think you've correctly used that formula. Applying it for $I_2$ gives us $$I_2=\int_{\pi/4}^{\pi/2}-\cos^{m-1}\left(\tfrac\pi 4-x\right)\sin^{n-1}\left(\tfrac\pi 4-x\right)\,dx$$ Instead, you can make the substitution $x\mapsto\frac\pi 2-x$ to obtain $$I_2=\int_0^{\pi/4}\cos^{m-1}(x)\sin^{n-1}(x)\,dx $$ which is similar to $I_1$.