Question Number 98151 by M±th+et+s last updated on 11/Jun/20
![∫e^(x^5 +8x^2 ) dx =((√π)/(4(√2)))e^x^5 erfi(2(√2)x)−((5(√π))/(4(128)(√2)))(super−erf_((hyper)) (2(√2)x))+c where[super−erf_((hyper)) (t)] is super−function in D_2 and [D_n ]](Q98151.png)
$$\int{e}^{{x}^{\mathrm{5}} +\mathrm{8}{x}^{\mathrm{2}} } {dx} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}\sqrt{\mathrm{2}}}{e}^{{x}^{\mathrm{5}} } {erfi}\left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right)−\frac{\mathrm{5}\sqrt{\pi}}{\mathrm{4}\left(\mathrm{128}\right)\sqrt{\mathrm{2}}}\left({super}−{erf}_{\left({hyper}\right)} \left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right)\right)+{c} \\ $$$$ \\ $$$${where}\left[{super}−{erf}_{\left({hyper}\right)} \left({t}\right)\right]\:{is}\:{super}−{function} \\ $$$${in}\:{D}_{\mathrm{2}} \:{and}\:\left[{D}_{{n}} \right] \\ $$
Answered by maths mind last updated on 12/Jun/20
$${i}\:{didnt}\:{find}\:{this}\:{super}−{erf}_{{hyoer}} \left({t}\right)\:\:{can}\:{you}\:{express}\:{definition} \\ $$$$ \\ $$
Commented by M±th+et+s last updated on 12/Jun/20
$${i}\:{don}'{t}\:{think}\:{we}\:{can}\:{solve}\:{this}\:{without} \\ $$$${using}\:{new}\:{special}\:{functions}\:{like}\:{this}\:{one} \\ $$$${i}\:{tried}\:{witb}\:{hyper}\:{geometric}\:{function} \\ $$$${and}\:{mejier}\:{G}−{function}\left({special}\:{high}\:{function}\right) \\ $$$${but}\:{i}\:{didn}'{t}\:{find}\:{any}\:{solutions}. \\ $$$${and}\:{i}\:{posted}\:{this}\:{question}\:{because}\:\:{mybe} \\ $$$${you}\:{have}\:{an}\:{idea}\:{to}\:{solve}\:{because}\:{you} \\ $$$${always}\:{do}. \\ $$$${i}\:{respect}\:{your}\:{mind}\:{sir} \\ $$$${to}\:{prof}.{math}\:{mind} \\ $$
Answered by M±th+et+s last updated on 12/Jun/20
![I=∫e^(x^5 +8x^2 ) dx ∫e^x^5 e^(8x^2 ) dx { ((u(x)=e^x^5 du=5x^4 e^x^5 )),((dv=e^(8x^2 ) v=(1/4)(√(π/2))erfi(2(√2)x))) :} where erfi(y) is the imaginary eror function I=((√π)/(4(√2))) e^x^5 erfi(2(√2)x)−((5π)/(4(√2)))∫x^4 e^x^5 erfi(2(√2)x)dx...✠ I_1 =∫x^4 e^x^5 erfi(2(√2)x)dx;let 2(√2)x=u dx=(1/(2(√2)))dy I_1 =(1/(128(√2)))∫y^4 e^((1/(128(√2)))y^5 ) erfi(y)dy=r∫y^4 e^(ry^5 ) erfi(y) dy ;(r=(1/(128(√2)))∈R) but erfi(y)=(i/(√π))Σ_(k=0) ^∞ (((−1)^(k−1) .H_(2k+i) (iy))/(2^(3k+(1/2)) .k!(2k+1)))and J=∫y^n .e^(ry) .erfi(y)dy J=((−n!)/(√π))(−r)^(−(n+1)) exp(((−r^2 )/4)).Σ_(m=0) ^n (((−r)^m )/(m!)).Σ_(k=0) ^m ((m),(k) )(((−r)/2))^(m−k) ((r/2)+y)^(k+1) .[(−((r/2)+y)^2 +y)^2 ]^2 .Γ(((k+1)/2);−((r/2)+y)^2 ] −r^(−(n+1)) .erfi(y)Γ(A+1,ny) n∈N so ⇒⇒Γ(n+1,ry);n∈N super integrative conversion T^(super−T^(−1) ) [∫y^4 e^(ry^5 ) erfi(y)dy]=d(super−erf_(hyper) (y)) then I_1 =r∫y^4 e^(ry^5 ) erfi(y)dy=r∫d(super−erf_(hyper) (y)) I_1 =r super−erf_((hyper)) (y)+c=(1/(128(√2)))super−erf_((hyper)) (2(√2)x)+c so I=((√π)/(4(√2)))e^x^5 erfi(2(√2)x)−((5(√π))/(4(128)(√2)))[super−erf_(hyper) (2(√2)x)]+c](Q98230.png)
$${I}=\int{e}^{{x}^{\mathrm{5}} +\mathrm{8}{x}^{\mathrm{2}} } {dx} \\ $$$$\int{e}^{{x}^{\mathrm{5}} } \:{e}^{\mathrm{8}{x}^{\mathrm{2}} } \:{dx\begin{cases}{{u}\left({x}\right)={e}^{{x}^{\mathrm{5}} } \:\:\:\:\:\:{du}=\mathrm{5}{x}^{\mathrm{4}} {e}^{{x}^{\mathrm{5}} } }\\{{dv}={e}^{\mathrm{8}{x}^{\mathrm{2}} } \:\:\:\:\:\:\:\:\:\:{v}=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\frac{\pi}{\mathrm{2}}}{erfi}\left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right)}\end{cases}} \\ $$$${where}\:{erfi}\left({y}\right)\:{is}\:{the}\:{imaginary}\:{eror}\:{function} \\ $$$$ \\ $$$${I}=\frac{\sqrt{\pi}}{\mathrm{4}\sqrt{\mathrm{2}}}\:{e}^{{x}^{\mathrm{5}} } {erfi}\left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right)−\frac{\mathrm{5}\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\int{x}^{\mathrm{4}} \:{e}^{{x}^{\mathrm{5}} } {erfi}\left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right){dx}...\maltese \\ $$$${I}_{\mathrm{1}} =\int{x}^{\mathrm{4}} {e}^{{x}^{\mathrm{5}} } {erfi}\left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right){dx};{let}\:\mathrm{2}\sqrt{\mathrm{2}}{x}={u}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{dy} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{128}\sqrt{\mathrm{2}}}\int{y}^{\mathrm{4}} {e}^{\frac{\mathrm{1}}{\mathrm{128}\sqrt{\mathrm{2}}}{y}^{\mathrm{5}} } {erfi}\left({y}\right){dy}={r}\int{y}^{\mathrm{4}} {e}^{{ry}^{\mathrm{5}} } {erfi}\left({y}\right)\:{dy}\:;\left({r}=\frac{\mathrm{1}}{\mathrm{128}\sqrt{\mathrm{2}}}\in\mathbb{R}\right) \\ $$$${but}\:{erfi}\left({y}\right)=\frac{{i}}{\sqrt{\pi}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} .{H}_{\mathrm{2}{k}+{i}} \left({iy}\right)}{\mathrm{2}^{\mathrm{3}{k}+\frac{\mathrm{1}}{\mathrm{2}}} .{k}!\left(\mathrm{2}{k}+\mathrm{1}\right)}{and}\:{J}=\int{y}^{{n}} .{e}^{{ry}} .{erfi}\left({y}\right){dy} \\ $$$${J}=\frac{−{n}!}{\sqrt{\pi}}\left(−{r}\right)^{−\left({n}+\mathrm{1}\right)} {exp}\left(\frac{−{r}^{\mathrm{2}} }{\mathrm{4}}\right).\underset{{m}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left(−{r}\right)^{{m}} }{{m}!}.\underset{{k}=\mathrm{0}} {\overset{{m}} {\sum}}\begin{pmatrix}{{m}}\\{{k}}\end{pmatrix}\left(\frac{−{r}}{\mathrm{2}}\right)^{{m}−{k}} \left(\frac{{r}}{\mathrm{2}}+{y}\right)^{{k}+\mathrm{1}} .\left[\left(−\left(\frac{{r}}{\mathrm{2}}+{y}\right)^{\mathrm{2}} +{y}\right)^{\mathrm{2}} \right]^{\mathrm{2}} .\Gamma\left(\frac{{k}+\mathrm{1}}{\mathrm{2}};−\left(\frac{{r}}{\mathrm{2}}+{y}\right)^{\mathrm{2}} \right] \\ $$$$−{r}^{−\left({n}+\mathrm{1}\right)} .{erfi}\left({y}\right)\Gamma\left({A}+\mathrm{1},{ny}\right)\:{n}\in{N}\:{so}\:\Rightarrow\Rightarrow\Gamma\left({n}+\mathrm{1},{ry}\right);{n}\in{N} \\ $$$$ \\ $$$${super}\:{integrative}\:{conversion}\:\overset{{super}−{T}^{−\mathrm{1}} } {{T}}\left[\int{y}^{\mathrm{4}} {e}^{{ry}^{\mathrm{5}} } {erfi}\left({y}\right){dy}\right]={d}\left({super}−{erf}_{{hyper}} \left({y}\right)\right) \\ $$$${then}\:{I}_{\mathrm{1}} ={r}\int{y}^{\mathrm{4}} {e}^{{ry}^{\mathrm{5}} } {erfi}\left({y}\right){dy}={r}\int{d}\left({super}−{erf}_{{hyper}} \left({y}\right)\right) \\ $$$${I}_{\mathrm{1}} ={r}\:{super}−{erf}_{\left({hyper}\right)} \left({y}\right)+{c}=\frac{\mathrm{1}}{\mathrm{128}\sqrt{\mathrm{2}}}{super}−{erf}_{\left({hyper}\right)} \left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right)+{c} \\ $$$${so}\: \\ $$$${I}=\frac{\sqrt{\pi}}{\mathrm{4}\sqrt{\mathrm{2}}}{e}^{{x}^{\mathrm{5}} } {erfi}\left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right)−\frac{\mathrm{5}\sqrt{\pi}}{\mathrm{4}\left(\mathrm{128}\right)\sqrt{\mathrm{2}}}\left[{super}−{erf}_{{hyper}} \left(\mathrm{2}\sqrt{\mathrm{2}}{x}\right)\right]+{c} \\ $$$$ \\ $$