Mixed ODE and FTC asked in 1985 in the Tokyo Metropolitan University-Science and Engineering

Exercise:

Consider a function $f$ defined on ${{\mathbb{R}}^{+}}$ to be $f\left( x \right)=\frac{2}{{{e}^{x}}-{{e}^{-x}}}\left( 1+\int_{1}^{x}{f\left( t \right)dt} \right)$

Find $\underset{\omega \to \infty }{\mathop{\lim }}\,\int_{\frac{1}{\omega }}^{\omega }{f\left( t \right)dt}$

Solution: Let $y=\int_{1}^{x}{f\left( t \right)dt}$ and we know that$\sinh x=\frac{{{e}^{x}}-{{e}^{-x}}}{2}$

then $f\left( x \right)=\frac{1}{\sinh x}\left( 1+y \right)$ by fundamental theorem of calculus $y'=f\left( x \right)$

so $y'=\frac{1+y}{\sinh x}\Leftrightarrow y'\sinh x=1+y\Leftrightarrow \frac{y'}{1+y}=\frac{1}{\sinh x}$

so $\int{\frac{d\left( 1+y \right)}{1+y}=\int{\frac{1}{\sinh x}dx}}\Leftrightarrow \ln \left| 1+y \right|=\int{\frac{dx}{\sinh x}}$

so $\frac{1}{\sinh x}=\frac{{{\cosh }^{2}}\left( \frac{x}{2} \right)-{{\sinh }^{2}}\left( \frac{x}{2} \right)}{2\sinh \left( \frac{x}{2} \right)\cosh \left( \frac{x}{2} \right)}=\frac{\cosh \left( \frac{x}{2} \right)}{\sinh \left( \frac{x}{2} \right)}-\frac{\sinh \left( \frac{x}{2} \right)}{\cosh \left( \frac{x}{2} \right)}$

so $\int{\frac{1}{\sinh x}dx}=\int{\frac{\cosh \left( \frac{x}{2} \right)}{\sinh \left( \frac{x}{2} \right)}dx}-\int{\frac{\sinh \left( \frac{x}{2} \right)}{\cosh \left( \frac{x}{2} \right)}dx}=\ln \left| \sinh \frac{x}{2} \right|-\ln \left| \cosh \frac{x}{2} \right|+c$

thus $\ln \left| 1+y \right|=\ln \left| \tanh \frac{x}{2} \right|+c\Leftrightarrow 1+y=c\tanh \frac{x}{2}\Leftrightarrow y\left( x \right)=c\tanh \frac{x}{2}-1$

Observe that when $x=1$ we get $y=\int_{1}^{1}{f\left( t \right)dt}=0$ i.e $y\left( 1 \right)=0$

$\Leftrightarrow y\left( 1 \right)=c\tanh \frac{1}{2}-1=0\Leftrightarrow c\tanh \frac{1}{2}=1\Leftrightarrow c=\coth \frac{1}{2}$ hence $f\left( x \right)=y'=\frac{1}{2}\coth \frac{1}{2}{{\operatorname{sech}}^{2}}\left( \frac{x}{2} \right)$

so $\underset{\omega \to \infty }{\mathop{\lim }}\,\int_{\frac{1}{w}}^{w}{f\left( t \right)dt}=\underset{\omega \to \infty }{\mathop{\lim }}\,\int_{\frac{1}{\omega }}^{\omega }{\frac{1}{2}\coth \left( \frac{1}{2} \right){{\operatorname{sech}}^{2}}\frac{t}{2}dt}$

                         $=\underset{\omega \to \infty }{\mathop{\lim }}\,\left[ \frac{1}{2}\coth \left( \frac{1}{2} \right)\left( 2\tanh \frac{\omega }{2}-2\tanh \frac{1}{2w} \right) \right]=\coth \frac{1}{2}\left( 1-0 \right)=\coth \frac{1}{2}$