How should I solve $ \int \frac{e^{2x-\frac {x^2}{4}}}{({x+4})^{8}}\, dx $?

Question

I have tried but nothing seems to work. I need help. $$ \int \frac{e^{2x-\frac{x^2}{4}}}{({x+4})^{8}}\, dx $$

if we get $$ t=x+4 $$ it yields:

$$ e^{-12}\int\frac{e^{4t-\frac{t^2}{4}}}{{t}^{8}}\, dt $$ Now how do we advance?

Answer 1

Too long for comments.

If you are concerned by $$I(a,b)=\int_0^\infty \frac {e^{-x^2}}{(x+a)^b}\,dx \quad \text{with}\quad a \geq 0\quad \text{and}\quad b \geq 0$$ we have $$I(a,b)=\frac{1}{2} \Gamma \left(\frac{1-b}{2}\right) \, _1F_1\left(\frac{b}{2};\frac{1}{2};-a^2\right)+a \Gamma \left(\frac{2-b}{2}\right) \, _1F_1\left(\frac{b+1}{2};\frac{3}{2};-a^2\right)-$$ $$\frac{a^{1-b}}{1-b}\, _2F_2\left(\frac{1}{2},1;\frac{2-b}{2},\frac{3-b}{2};-a^2\right)$$ where appears twice the Kummer confluent hypergeometric function and a generalized hypergeometric function.

Answer 2

$$\int_{}^{}\frac{e^{2x-\frac{x^2}{4}}}{(x+4)^8}dx$$ After doing some efforts one can complete the square in the power of $e$ as follows, $$e^{4}2^{-7}\int_{}^{}\frac{e^{-(u+2)^2} e^{4u}}{(u+2)^8}du$$ Now let's substitute $t=u+2 \implies dt=du$. $$e^{3}2^{-7}\int_{}^{}\frac{e^{-(t)^2} e^{4t}}{(t)^8}du \rightarrow (1)$$ Now I will calculate the integral, $$\int_{}^{}\frac{e^{-t^{2}} e^{4t}}{t^8}dt$$ Let, $$ I(a)=\int_{}^{}\frac{e^{at^2} e^4t}{t^8} dt$$ Notice that, $$ I^{(4)}(a)=\int_{}^{}\frac{e^{at^2} t^8 e^4t}{t^8}dt$$ $$ I^{(4)}(a)=\int_{}^{} e^{at^2} e^4tdt$$ $$ I^{(4)}(a)=\int_{}^{} e^{at^2+4t} dt$$ To express this in terms of error function. Notice that, $$ f(x)=ax^2+bx+c=a(x-p)^2+k$$ Where, $p=\frac{-b}{2a}, k=ap^2+bp+c$. This idea is analogous to conversion of cubic to a depressed one. $$ I^{(4)}(a)=e^{k}\int_{}^{} e^{a(t-p)^2} dt $$ Put, $s=(t-p)\sqrt{a} \implies dt=\frac{1}{\sqrt{a}}ds$ $$ I^{(4)}(a)=\frac{e^k}{\sqrt{a}}\int_{}^{} e^{s^2}ds$$ $$ I^{(4)}(a)=\frac{e^k}{\sqrt{a}} erfi(s)$$ $$ I^{(4)}(a)=\frac{e^k}{\sqrt{a}} erfi((t-p)\sqrt{a})$$

Now, one should integrate on both sides $4$ times to get $I(a)$. Evaluating $I(-1)$ and placing it in $eq.(1)$ one can evaluate the Integral in terms of error function. As the process suggests, the anti derivative in terms of error functions is complicated. The answer by @Claude Leibovici in terms of hypergeometric functions is best.

Answer 3

The function $$ f(y)=\int_a^y \frac{e^{2x-\frac{x^2}{4}}}{(x+4)^8}dx $$ Is not defined if $a$ and $y$ are on opposite sides of $-4$. This is so because the integrand diverges at $x=4$ and the integral cannot be computed in this case. When $y$ is not too large, for example if $|y|<8$ (the number $8$ is taken since the integrand is already very small there) the integrand is essentially zero and do not contribute to the integral as can be easily seen by plugging in some numbers. For $|y|<8$ and $a>4$ we can easily get a good approximation for the integral. For $0<y<8$ we write: $$ \frac1{(x+4)^8} =e^{-8\ln(x+4)}\approx e^{-8(\ln 4+\frac {x}{4}-\frac {x^2}{32} +\frac{x^3}{192})} $$ Subtituting this into the integral we get: $$ f(y)\approx \frac 1{4^8}\int_a^y e^{-\frac{x^3}{24}}dx=\frac{(24)^{ 1/3}}{3\cdot4^8} \Gamma[(\frac13,\frac{y^3}{24})-\Gamma(\frac13,\frac{a^3}{24})] $$ where the last step was performed with the help of Wolfram Alpha and contains the incomplete Gamma functionץ For $y$ greater than $8$ the integrand is very small and do not contribute to the integral, so we can take the integral at that range as essentially constant.