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Question Number 205396 by MathedUp last updated on 20/Mar/24
Can′t understand why  ∫_0 ^∞ [e_  ^(sin(z)) −e^(cos(z)) ]dz isn′t (1/2)π(I_0 ^  (1)−L_0 (1))≈3.104...  I Just guessing f(0)= (1/2)πI_0 (1)−L_0 (1)  not lim_(s→0^+ ) f(s)  ex.find Value lim_(z→0)  g(z)    g(z)=((sin(z))/z)  , g(0)=3  lim_(z→0) g(z)=1 ,  But g(0)=3 ∴ g(0)≠lim_(z→0) g(z)  function f(s)=∫_0 ^∞  e^(−st+sin(t)) dt−∫_0 ^∞  e^(−st+cos(t)) dt  I_ν (z) Modified Bessel Function  L_ν (z) Modified Struve Function  ∫_0 ^∞  e^(−st+sin(t)) dt=  (1/s)+Σ_(h=1) ^∞  Π_(k=1) ^h  (1/(s^2 +(2k−1)^2 ))+(1/s) Σ_(h=1) ^∞  Π_(k=1) ^h  (1/(s^2 +4k^2 ))  ∫_0 ^∞  e^(−st+cos(t)) dt=  e^((π/2)s) ∫_0 ^∞  e^(−st+sin(t)) dt−e^((π/2)s) ∫_0 ^(π/2)  e^(−st+sin(t)) dt
$$\mathrm{Can}'\mathrm{t}\:\mathrm{understand}\:\mathrm{why} \\ $$$$\int_{\mathrm{0}} ^{\infty} \left[{e}_{\:} ^{\mathrm{sin}\left({z}\right)} −{e}^{\mathrm{cos}\left({z}\right)} \right]\mathrm{d}{z}\:\mathrm{isn}'{t}\:\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\boldsymbol{\mathrm{I}}_{\mathrm{0}} ^{\:} \left(\mathrm{1}\right)−\boldsymbol{\mathrm{L}}_{\mathrm{0}} \left(\mathrm{1}\right)\right)\approx\mathrm{3}.\mathrm{104}… \\ $$$$\mathrm{I}\:\mathrm{Just}\:\mathrm{guessing}\:{f}\left(\mathrm{0}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\pi\boldsymbol{\mathrm{I}}_{\mathrm{0}} \left(\mathrm{1}\right)−\boldsymbol{\mathrm{L}}_{\mathrm{0}} \left(\mathrm{1}\right) \\ $$$$\mathrm{not}\:\underset{{s}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{f}\left({s}\right) \\ $$$$\mathrm{ex}.\mathrm{find}\:\mathrm{Value}\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{g}\left({z}\right)\: \\ $$$$\:\mathrm{g}\left({z}\right)=\frac{\mathrm{sin}\left({z}\right)}{{z}}\:\:,\:\mathrm{g}\left(\mathrm{0}\right)=\mathrm{3} \\ $$$$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}g}\left({z}\right)=\mathrm{1}\:,\:\:\mathrm{But}\:\mathrm{g}\left(\mathrm{0}\right)=\mathrm{3}\:\therefore\:\mathrm{g}\left(\mathrm{0}\right)\neq\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}g}\left({z}\right) \\ $$$$\mathrm{function}\:{f}\left({s}\right)=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{st}+\mathrm{sin}\left({t}\right)} \mathrm{d}{t}−\int_{\mathrm{0}} ^{\infty} \:{e}^{−{st}+\mathrm{cos}\left({t}\right)} \mathrm{d}{t} \\ $$$$\boldsymbol{\mathrm{I}}_{\nu} \left({z}\right)\:\mathrm{Modified}\:\mathrm{Bessel}\:\mathrm{Function} \\ $$$$\boldsymbol{\mathrm{L}}_{\nu} \left({z}\right)\:\mathrm{Modified}\:\mathrm{Struve}\:\mathrm{Function} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{st}+\mathrm{sin}\left({t}\right)} \mathrm{d}{t}= \\ $$$$\frac{\mathrm{1}}{{s}}+\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{{k}=\mathrm{1}} {\overset{{h}} {\prod}}\:\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{{s}}\:\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{{k}=\mathrm{1}} {\overset{{h}} {\prod}}\:\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{4}{k}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{st}+\mathrm{cos}\left({t}\right)} \mathrm{d}{t}= \\ $$$${e}^{\frac{\pi}{\mathrm{2}}{s}} \int_{\mathrm{0}} ^{\infty} \:{e}^{−{st}+\mathrm{sin}\left({t}\right)} \mathrm{d}{t}−{e}^{\frac{\pi}{\mathrm{2}}{s}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{e}^{−{st}+\mathrm{sin}\left({t}\right)} \mathrm{d}{t} \\ $$

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