Question Number 220015 by SdC355 last updated on 04/May/25
![∫_0 ^( ∞) ∣∣J_ν (r)∣∣e^(−rt) dr=Σ_(h=1) ^∞ ∫_z_h ^( z_(h+1) ) J_ν (r)e^(−rt) dr z_j is point of J_ν (z)=0 , z_1 =0 Σ_(h=1) ^∞ [F(r,t)]_(r=z_h ) ^(r=z_(h+1) ) i can′t solve anymore](Q220015.png)
$$\int_{\mathrm{0}} ^{\:\infty} \:\mid\mid{J}_{\nu} \left({r}\right)\mid\mid{e}^{−{rt}} \:\mathrm{d}{r}=\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\int_{{z}_{{h}} } ^{\:{z}_{{h}+\mathrm{1}} } \:{J}_{\nu} \left({r}\right){e}^{−{rt}} \mathrm{d}{r} \\ $$$${z}_{{j}} \:\mathrm{is}\:\mathrm{point}\:\mathrm{of}\:\:{J}_{\nu} \left({z}\right)=\mathrm{0}\:,\:{z}_{\mathrm{1}} =\mathrm{0} \\ $$$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left[{F}\left({r},{t}\right)\right]_{{r}={z}_{{h}} } ^{{r}={z}_{{h}+\mathrm{1}} } \: \\ $$$$\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{anymore} \\ $$
Answered by MrGaster last updated on 04/May/25
![Σ_(h=1) ^∞ [−((J_ν (r)e^(−rt) )/t)]_(r=z_h ) ^(r=z_(h+1) ) =((J_ν (0))/t)+lim_(r→∞) ((J_ν (r)e^(−rt) )/t) −((J_ν (0))/t) −((J_ν (0))/t)=0 for ν>0 lim_(r→∞) ((J_ν (r)e^(−rt) )/t)=0 ∫_0 ^∞ J_ν (r)e^(−rt) dr=((((√(t^2 +1))−t)^ν )/( (√(t^2 +1))))](Q220024.png)
$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\left[−\frac{{J}_{\nu} \left({r}\right){e}^{−{rt}} }{{t}}\right]_{{r}={z}_{{h}} } ^{{r}={z}_{{h}+\mathrm{1}} } =\frac{{J}_{\nu} \left(\mathrm{0}\right)}{{t}}+\underset{{r}\rightarrow\infty} {\mathrm{lim}}\frac{{J}_{\nu} \left({r}\right){e}^{−{rt}} }{{t}} \\ $$$$\cancel{−\frac{{J}_{\nu} \left(\mathrm{0}\right)}{{t}}} \\ $$$$−\frac{{J}_{\nu} \left(\mathrm{0}\right)}{{t}}=\mathrm{0}\:\mathrm{for}\:\nu>\mathrm{0} \\ $$$$\underset{{r}\rightarrow\infty} {\mathrm{lim}}\frac{{J}_{\nu} \left({r}\right){e}^{−{rt}} }{{t}}=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\infty} {J}_{\nu} \left({r}\right){e}^{−{rt}} {dr}=\frac{\left(\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}−{t}\right)^{\nu} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$
Commented by MrGaster last updated on 04/May/25
Note: The original absolute value is unnecessary, so it is removed.