Is there a closed form solution for the following integral
$$\int_0^z\frac{1}{1+z-x}\frac{1}{(1+x)^K}\,dx$$
for a positive integer $K\geq 1$, and positive real number $z\geq 0$? I searched the table of integrals, but couldn't find something similar.
Closed-form solution for an integral
calculusclosed-formintegration
Related Solutions
The result reads: \begin{eqnarray} &&\int\limits_0^\infty x e^{-a x^2-b x} \text{erf}(c x+d) dx= \frac{e^{\frac{b^2}{4 a}}}{4 a} \left( \right.\\ && \frac{2 b \left(\arctan\left(\frac{\sqrt{a} (b+2 c d)}{2 a d-b c}\right)+\arctan\left(\frac{2 \sqrt{a} d}{b}\right)-\arctan\left(\frac{c}{\sqrt{a}}\right)\right)}{\sqrt{\pi } \sqrt{a}} +\\ && 2 e^{-\frac{b^2}{4 a}} \text{erf}(d)-\frac{\sqrt{\pi } b \text{erf}\left(\frac{2 a d-b c}{2 \sqrt{a \left(a+c^2\right)}}\right)}{\sqrt{a}}+\frac{2 c e^{-\frac{(b c-2 a d)^2}{4 a \left(a+c^2\right)}} \text{erfc}\left(\frac{b+2 c d}{2 \sqrt{a+c^2}}\right)}{\sqrt{a+c^2}} +\\ && -\frac{4 \sqrt{\pi } b \left(T\left(\frac{2 a d-b c}{\sqrt{2} \sqrt{a \left(a+c^2\right)}},\frac{\sqrt{a} (b+2 c d)}{2 a d-b c}\right)+T\left(\frac{b}{\sqrt{2} \sqrt{a}},\frac{2 \sqrt{a} d}{b}\right)\right)}{\sqrt{a}}\\ &&\left.\right) \end{eqnarray} where $T(\cdot,\cdot)$ is the Owen's t function https://en.wikipedia.org/wiki/Owen%27s_T_function .
For[count = 1, count <= 200, count++,
{a, b, c} = RandomReal[{0, 5}, 3, WorkingPrecision -> 50];
d = RandomReal[{-5, 5}, WorkingPrecision -> 50];
I1 = NIntegrate[x Exp[-a x^2 - b x] Erf[c x + d], {x, 0, Infinity},
WorkingPrecision -> 20];
1/a NIntegrate[
x Exp[-x^2 - b/Sqrt[a] x] Erf[c/Sqrt[a] x + d], {x, 0, Infinity}];
1/a Exp[
b^2/(4 a)] NIntegrate[(x - b/(2 Sqrt[a])) Exp[-x^2 ] Erf[
c/Sqrt[a] ( x - b/(2 Sqrt[a])) + d], {x, b/(2 Sqrt[a]),
Infinity}];
1/a Exp[
b^2/(4 a)] (NIntegrate[(x) Exp[-x^2 ] Erf[
c/Sqrt[a] ( x - b/(2 Sqrt[a])) + d], {x, b/(2 Sqrt[a]),
Infinity}] -
b Sqrt[Pi]/(2 Sqrt[a]) 2 T[b/Sqrt[2 a],
c/Sqrt[a], (-((b c)/(2 a)) + d) Sqrt[2]]);
I2 = E^(b^2/(4 a))/(
4 a) ((2 b)/(
Sqrt[a] Sqrt[\[Pi]]) (-ArcTan[c/Sqrt[a]] +
ArcTan[(2 Sqrt[a] d)/b] +
ArcTan[(Sqrt[a] (b + 2 c d))/(-b c + 2 a d)]) +
2 E^(-(b^2/(4 a))) Erf[d] - (b Sqrt[\[Pi]])/Sqrt[a]
Erf[(-b c + 2 a d)/(2 Sqrt[a (a + c^2)])] + (2 c)/Sqrt[
a + c^2] E^(-((b c - 2 a d)^2/(4 a (a + c^2))))
Erfc[(b + 2 c d)/(2 Sqrt[a + c^2])] - (4 b Sqrt[\[Pi]])/Sqrt[
a] (OwenT[b/(Sqrt[2] Sqrt[a]), (2 Sqrt[a] d)/b] +
OwenT[(-b c + 2 a d)/(Sqrt[2] Sqrt[a (a + c^2)]), (
Sqrt[a] (b + 2 c d))/(-b c + 2 a d)]));
If[Abs[I2/I1 - 1] > 10^(-3),
Print["results do not match", {a1, a2, c, d, {I1, I2}}]; Break[]];
If[Mod[count, 10] == 0, PrintTemporary[count]];
];
Best Answer
Hint:
Let $$ I(k)=\frac {1}{z+2} \int_0^z \frac {(1+x)+(1+z-x)}{(1+z-x)(1+x)^k} dx $$
$$I(k)=\frac {1}{z+2} \int_0^z \frac{1}{(1+z-x)(1+x)^{k-1}}+\frac {1}{z+2}\int_0^z (1+x)^{-k}$$
Therefore $$I(k)=\frac {I(k-1)}{z+2} +\frac {(1+z)^{1-k}-1}{(1-k)(z+2)}$$
If you now want then you can solve this recursion but I don't think that would be a great and easy job to do.
Hope it helps