Let ${f}({x})=\displaystyle \int_{{x}}^{{x}+1} \sin \left({e}^{{t}}\right) {dt}$ then:
(A) $\displaystyle \int_{0}^{\infty} f(x) d x \leq 2 $
(B) $\displaystyle \int_{0}^{\infty} f(x) d x > -e $
(C) $f(x)$ has local maxima or global maxima at ${x}=\ln \left(\frac{\pi}{1+\mathrm{e}}\right)$
(D)$f(x)$ is continuous

By Leibniz integral rule: $$f'(x)=\sin(e^{x+1})-\sin(e^x)$$ For maxima: $$f'(x)=0\\\sin(e^{x+1})-\sin(e^x)=0\\e^{x+1}=n\pi+(-1)^ne^x$$ For $n=1$: $${x}=\ln \left(\frac{\pi}{1+\mathrm{e}}\right)\\f''(x)=\mathrm{e}^{x+1}\cos\left(\mathrm{e}^{x+1}\right)-{e}^x\cos\left({e}^x\right)$$ And using calculator(I don't know how to approximate it without it): $$f''\left(\ln \left(\frac{\pi}{1+\mathrm{e}}\right)\right)<0$$ So (C) is correct.

For (A) and (B) : I tried substituting $u=e^t$ $$f(x)=\int_{e^x}^{e^{x+1}}\dfrac{\sin u}{u} du$$ And integrating by parts(taking Ist as $\dfrac1u$ and 2nd as $\sin(u)$ gives: $$f(x)=\left[\dfrac{-\cos u}{u}\right]_{e^x}^{e^{x+1}} - \int_{e^x}^{e^{x+1}} \dfrac{\cos u}{u^2}du $$ which I am unable to simplify.

The given answer is



3 Answers 3


Clearly $$f(x)=-e^{-x-1}\cos e^{x+1}+e^{-x}\cos e^{x} - \int_{e^x}^{e^{x+1}} \dfrac{\cos u}{u^2}du $$ and hence $$|f(x)|\le e^{-x}+e^{-x-1}+\int_{e^x}^{e^{x+1}} \dfrac{1}{u^2}du =\dfrac{2}{e^x}. $$ So $$\int_{0}^{\infty} f(x) d x \le \int_{0}^{\infty} |f(x)| d x=\int_{0}^{\infty} e^x d x=2. $$ On the other hand, using $$ \cos x\ge -1, -\cos x\ge -1 $$ and $$ \int_a^bf(x)dx\ge -\int_a^b|f(x)|dx$$ you have $$f(x)\ge-e^{-x-1}-e^{-x} - \int_{e^x}^{e^{x+1}} \bigg|\dfrac{1}{u^2}\bigg|du=-2e^{-x-1} $$ and hence $$\int_{0}^{\infty} f(x) d x \ge\int_{0}^{\infty} (-2e^{-x-1}) d x=-2e^{-1}>-e. $$ Now you can have (A) and (B).

  • $\begingroup$ Can you also comment on D part? $\endgroup$
    – Wolgwang
    Commented Apr 22, 2023 at 0:26
  • 1
    $\begingroup$ Since $\sin(e^t)$ is continuous, so is $f(x)$ by using properties of Riemann integrals. $\endgroup$
    – xpaul
    Commented Apr 22, 2023 at 2:55

Here is a graph of the function $f(x)$

enter image description here

The integral over $f$ which is defined by

$$g=\int_0^{\infty}\int_{{x}}^{{x}+1} \sin \left({e}^{{t}}\right) {dt} \ dx= \int_0^{\infty}(\text{Si}\left(e^{x+1}\right)-\text{Si}\left(e^x\right))\;dx$$

can be calculated explicitly as follows:

Decomposing the integration region $x$ into intervals of length $1$ we see that the integral is telescoping:

$$g=\lim_{n\to\infty}\int_0^{n}(\text{Si}\left(e^{x+1}\right)-\text{Si}\left(e^x\right))\;dx\\ =\int_0^{1}(\text{Si}\left(e^{x+1}\right)-\text{Si}\left(e^x\right))\;dx+\int_1^{2}(\text{Si}\left(e^{x+1}\right)-\text{Si}\left(e^x\right))\;dx+ ... \\ =\int_0^{1}(\color{blue}{\text{Si}\left(e^{x+1}\right)}-\text{Si}\left(e^x\right))\;dx\overset{x\to x+1}+\int_{0}^{1}(\text{Si}\left(e^{x+2}\right)\color{blue}{-\text{Si}\left(e^{x+1}\right)})\;dx+ ...\\ =\int_0^{1}(-\text{Si}\left(e^x\right))\;dx+\int_{0}^{1}\text{Si}\left(e^{x+2}\right)\;dx+ ... \\= \int_0^{1}(-\text{Si}\left(e^x\right))\;dx+\lim_{n\to\infty}\int_{0}^{1}\text{Si}\left(e^{x+n}\right)\;dx\\ $$

Since $\lim_{n\to\infty}\text{Si}\left(e^{x+n}\right)) =\frac{\pi}{2}$ we get

$$g=\frac{\pi }{2}-\int_0^1 \text{Si}\left(e^x\right) \, dx$$


$$h=\int_0^1 \text{Si}\left(e^x\right) \, dx \overset{x\to \log(s)}= \int_1^e \frac{\text{Si}\left(s\right)}{s} \, ds\overset{IBP}= \text{Si}\left(e\right)- \int_1^e \text{sinc}(s) \log (s) \, ds$$

which can be written, ($\text{Si}\left(e\right)$ cancels out), in terms of hypergeometric functions so that the question in the heading of the OP has the following answer:

$$g=\frac{\pi }{2}+\, _2F_3\left(\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2};-\frac{1}{4}\right)-e \, _2F_3\left(\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2};-\frac{e^2}{4}\right)\simeq 0.160985$$

Bonus: we can also integrate $f$ over the whole $x$-axis. The $\text{Si}$-terms cancel out and the result is just $\frac{\pi}{2}$


A slightly alternate solution which I came up with the help of Ted Shifrin .

Changing the order of integration of $\displaystyle \int_{0}^{\infty} f(x) d x \leq 2 $ gives: $$\int_0^{\infty}\int_{{x}}^{{x}+1} \sin \left({e}^{{t}}\right) {dt} \ dx=\int_0^{1}\int_{0}^{t} \sin \left({e}^{{t}}\right) {dx} \ dt+ \int_1^{\infty}\int_{t-1}^{t} \sin \left({e}^{{t}}\right) {dx} \ dt$$ Let $$I_1=\int_0^{1}\int_{0}^{t} \sin \left({e}^{{t}}\right) {dx} \ dt=\int_0^1t\sin(e^t)dt\\ I_2=\int_1^{\infty}\int_{t-1}^{t} \sin \left({e}^{{t}}\right) {dx} \ dt=\int_1^{\infty}\sin(e^t)dt$$

Now for $I_1$ : $$\int_0^1t\sin(e^t)dt\le\int_0^1tdt=\dfrac12<2$$

for $I_2$ , from integration by parts: $$I_2=\left[\dfrac{-\cos u}{u}\right]_{e}^{e^{\infty}} - \int_{e}^{e^{\infty}} \dfrac{\cos u}{u^2}du\\ =\dfrac{\cos(e)}{e}-\int_{e}^{e^{\infty}} \dfrac{\cos u}{u^2}du\\ I_2\ge\dfrac{\cos(e)}{e}+\int_{e}^{e^{\infty}} \dfrac{1}{u^2}du=\dfrac{\cos(e)-1}{e}\approx -0.70 $$ Also,$$I_2<\dfrac{\cos(e)+1}{e}\approx=0.03$$

So (A) and (B) are correct


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