Proof of $\int_0^{a}{\rm erf}(x)e^{x^2}(a^2-x^2)^{-1/2}xdx=\sqrt{\pi}(e^{a^2}-1)/2$

Question

This question comes from A table of integrals of the Error functions, one of the special definite integral is given by $$ \int_0^{a}{\rm erf}(x)e^{x^2}(a^2-x^2)^{-1/2}x\ dx=\frac{\sqrt{\pi}}{2}(e^{a^2}-1),\ \ a>0, $$ where ${\rm erf}(x)$ is the error function. How can I prove this result? I've tried mathematica but got nothing.

Answer 1

$$ \text{erf}(x)\,e^{x^2}\,x=\sum_{n=1}^\infty \frac{x^{2n}}{\Gamma \left(\frac{2 n+1}{2}\right)} $$ $$\int_0^a \frac {x^{2n}}{\sqrt{a^2-x^2}}\,dx= a^{2 n} \frac{\sqrt{\pi }\,\,\Gamma \left(\frac{2n+1}{2}\right)}{2 \Gamma (n+1)}$$

So $$\int_0^a \frac {\text{erf}(x)\,e^{x^2}\,x}{\sqrt{a^2-x^2}}\,dx=\sum_{n=1}^\infty \frac{\sqrt{\pi }\, a^{2 n}}{2 \,n!}=\frac{\sqrt{\pi }}{2} \left(e^{a^2}-1\right)$$

Edit

As you wrote in comments $$\int_0^{b<a} \frac {\text{erf}(x)\,e^{x^2}\,x}{\sqrt{a^2-x^2}}\,dx= \frac{1}{2}\sum _{n=1}^{\infty } a^{2 n} \frac{ B_{\frac{b^2}{a^2}}\left(n+\frac{1}{2},\frac{1}{2}\right)}{ \Gamma \left(\frac{2 n+1}{2} \right)}$$

Update

Using the same procedure $$I_k=\int_0^a \frac {\text{erf}(x)\,e^{x^2}\,x^{2k+1}}{\sqrt{a^2-x^2}}\,dx$$ $$I_k=a^{2(k+1)}\,\frac{\Gamma \left(k+\frac{3}{2}\right)}{\Gamma (k+2)}\,\, _2F_2\left(1,k+\frac{3}{2};\frac{3}{2},k+2;a^2\right)$$ which write $$I_k=\frac{\sqrt \pi } { \alpha_k}\big(P_k \,e^{a^2}-Q_k\Big)$$ where the first $\alpha_k$

$$\{2,4,16,32,256,512,2048,4096,65536,131072,524288,\cdots\}$$ correspond to sequence $A101926$ in $OEIS$

$$\left( \begin{array}{cc} k & P_k \\ 0 & 1 \\ 1 & 2 a^2-1 \\ 2 & 8 a^4-8 a^2+6 \\ 3 & 16 a^6-24 a^4+36 a^2-30 \\ 4 & 128 a^8-256 a^6+576 a^4-960 a^2+840 \\ 5 & 256 a^{10}-640 a^8+1920 a^6-4800 a^4+8400 a^2-7560 \\ \end{array} \right)$$

$$\left( \begin{array}{cc} k & Q_k \\ 0 & 1 \\ 1 & a^2-1 \\ 2 & 3 a^4-2 a^2+6 \\ 3 & 5 a^6-3 a^4+6 a^2-30 \\ 4 & 35 a^8-20 a^6+36 a^4-120 a^2+840 \\ 5 & 63 a^{10}-35 a^8+60 a^6-180 a^4+840 a^2-7560 \\ \end{array} \right)$$