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Question Number 77752 by abdomathmax last updated on 09/Jan/20

$$letf\left(\lambda \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin(\lambda e^x+e^{-x})}{x^2+{\lambda }^2}dxwith\lambda ⩾0$$$$1)detdrmineaexplicitformoff\left(\lambda \right)$$$$2)calculate{f}^{{}^{\prime }}\left(\lambda \right)atformofintergralandfind$$$$itsvalue.$$

Commented by mathmax by abdo last updated on 10/Jan/20

$$f\left(\lambda \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin(\lambda e^x+e^{-x})}{x^2+{\lambda }^2}\Rightarrow f\left(\lambda \right){=}_{x=\lambda t}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin(\lambda e^{\lambda t}+e^{-\lambda t})}{{\lambda }^2(t^2+1)}\lambda \left(dt\right)$$$$=\frac{1}{\lambda }{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin(\lambda e^{\lambda t}+e^{-\lambda t})}{t^2+1}dt(wetake\lambda >0)\Rightarrow$$$$\lambda f\left(\lambda \right)=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i(\lambda e^{\lambda t}+e^{-\lambda t})}}{t^2+1}dt\right)letW\left(z\right)=\frac{e^{i(\lambda e^{\lambda z}+e^{-\lambda z})}}{z^2+1}$$$$\Rightarrow W\left(z\right)=\frac{e^{i(\lambda e^{\lambda z}+e^{-\lambda z})}}{(z-i)(z+i)}and{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi Res(W,i)$$$$=2i\pi \times \frac{e^{i(\lambda e^{i\lambda }+e^{-i\lambda })}}{2i}wehave$$$$i(\lambda e^{i\lambda }+e^{-i\lambda })=i(\lambda cos\left(\lambda \right)+\lambda isin\left(\lambda \right)+cos\left(\lambda \right)-isin\left(\lambda \right))$$$$=i\lambda cos\left(\lambda \right)-\lambda sin\left(\lambda \right)+icos\left(\lambda \right)+sin\lambda$$$$=(1-\lambda )sin\lambda +i(1+\lambda )cos\left(\lambda \right)\Rightarrow$$$$e^{i(\lambda e^{i\lambda }+e^{-i\lambda )}}=e^{(1-\lambda )sin\left(\lambda \right)}\{cos(1+\lambda )cos\lambda +isin(1+\lambda )cos\lambda \}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=\pi e^{(1-\lambda )sin\lambda }cos\left((1+\lambda )cos\lambda \right)+i\pi e^{(1-\lambda )sin\lambda }sin\left((1+\lambda )cos\lambda \right)$$$$\Rightarrow \lambda f\left(\lambda \right)=\pi e^{(1-\lambda )sin\lambda }sin\left\{(1+\lambda )cos\lambda \right\}\Rightarrow$$$$f\left(\lambda \right)=\frac{\pi }{\lambda }{e}^{(1-\lambda )sin\lambda }sin\left\{(1+\lambda )cos\lambda \right\}(\lambda >0)$$

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