Is $I(t)=\int_{0}^{\pi/2} (x-\pi/2)\exp\bigl(-t\sec(x)\bigr) dx$ continuous?

Question

$$I(t)=\int_{0}^{\pi/2}(x-\pi/2)\exp\bigl(-t\sec(x)\bigr) dx$$ with $t>0$

The original problem that I had was much simpler, but I am curious about this function, is it continuous or not? I don't have any idea about how to demonstrate its continuity. Is it?

Answer 1

Yes, $I(t)$ is continuous. To prove so, you can use the dominated convergence theorem and majorize the integrand by the constant function $$ x \mapsto \frac{\pi}{2}. $$

In case you want to avoid the DCT, you can argue as follows: let $\varepsilon > 0$ and let $t > 0$ be fixed. We want to show that there is $\delta > 0$ so that if $|h| < \delta$, then $|I(t+h) - I(t)| < \varepsilon$. To make $|I(t+h) - I(t)|$ small, we begin by writing $$ \begin{split} |I(t+h) - I(t)| &= \left| \int_{0}^{\frac{\pi}{2}} \left( x - \frac{\pi}{2} \right)\left[ \exp(-(t+h) \sec x )- \exp(-t \sec x) \right] \,dx \right| \\ &\leq \int_{0}^{\frac{\pi}{2} - \frac{\varepsilon}{ \pi} }\left| x - \frac{\pi}{2} \right| \left| \exp(-t \sec x)( \exp(-h \sec x) - 1) \right| \, dx + \int_{\frac{\pi}{2} - \frac{\varepsilon}{ \pi} }^{\frac{\pi}{2}} \left| x - \frac{\pi}{2} \right| \cdot 1 \, dx \\ &\leq \frac{\varepsilon}{2}+ \frac{\pi}{2} \int_{0}^{\frac{\pi}{2} - \frac{\varepsilon}{ \pi} } \left | \exp(-h \sec x) - 1 \right| \, dx. \end{split} $$ Since the exponential function is Lipschitz on compact intervals, there exists $C$ so that $|\exp(u)-\exp(v)| \leq C|u-v|$ for any $u,v \in \left[- \sec(\frac{\pi}{4} - \frac{\varepsilon}{ \pi}), \sec(\frac{\pi}{4} - \frac{\varepsilon}{ \pi}) \right]$. If we choose $$ \delta < \frac{2 \varepsilon}{\pi^2 C \sec\left(\frac{\pi}{4} - \frac{ \varepsilon}{\pi} \right)} $$ and assume that $|h| < \delta$, then $|-h \sec x| < \frac{2 \varepsilon}{\pi^2 C}$, so $$ \int_{0}^{\frac{\pi}{2} - \frac{\varepsilon}{ \pi} } \left | \exp(-h \sec x) - 1 \right| \, dx \leq \int_{0}^{\frac{\pi}{2} - \frac{\varepsilon}{ \pi} } C \left( \frac{2 \varepsilon}{\pi^2 C} \right) \, dx \leq \frac{\varepsilon}{\pi}, $$ which means that $$ |I(t+h)-I(t)| \leq \varepsilon. $$

Answer 2

Just for your curiosity since you already received a good answer.

We can have a quite good approximation of $I(t)$ using Taylor series built at $x=0$. This would give $$e^t\,{e^{-t \sec (x)}}=1-\frac{t }{2}x^2+\frac{ t (3 t-5) }{24}x^4-\frac{t(15 t^2-75 t+61)}{720} x^6+O\left(x^8\right)$$ Integrating the whole expression and using the bounds, this would give as an approximation $$I(t)\sim\frac{\pi ^2 \,e^{-t}}{10321920}(15 \pi ^6 t^3-\left(672 \pi ^4+75 \pi ^6\right) t^2+\left(26880 \pi ^2+1120 \pi ^4+61 \pi ^6\right) t-1290240)$$ which is not "too" bad when compared with numerical integration.