The Hermite polynomials $H_n(x)$ have the following explicit expression:
$$H_n(x)=\sum_{m=0}^{\lfloor n/2\rfloor}\frac{(-1)^mn!2^{n-2m}}{m!(n-2m)!}\,x^{n-2m}\,.$$
On the other hand, the Legendre polynomials have the following explicit expression:
$$P_n(x)=\frac1{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}\,,$$
which implies
$$t^{n+1}P_n(x/t)=\frac1{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}\,t^{2k+1}\,.$$
For $k\geq0$, let $A_k=\int_x^\infty e^{-t^2}t^{2k+1}\,dt$. Integrating by parts we obtain
$$\begin{align*}
A_k=&\,\int_x^\infty\frac{t^{2k}}{-2}e^{-t^2}(-2t)\,dt\\[2mm]
=&\,\frac{t^{2k}e^{-t^2}}{-2}\biggl|_{t=x}^{t=\infty}-\int_x^\infty-ke^{-t^2}t^{2k-1}\,dt\\[2mm]
=&\,\frac{x^{2k}e^{-x^2}}2+kA_{k-1}\,;
\end{align*}$$
in particular $A_0=e^{-x^2}/2$, and so for each $k\geq0$ we have
$$\sum_{r=1}^k\frac{A_r-rA_{r-1}}{r!}=\sum_{r=1}^k\frac{x^{2r}e^{-x^2}}{2\cdot r!}$$
that is
$$\frac{A_k}{k!}-\frac{A_0}{0!}=\frac{e^{-x^2}}2\sum_{r=1}^k\frac{x^{2r}}{r!}\,.$$
Therefore we have, for $k\geq0$:
$$A_k=\frac{e^{-x^2}}2\sum_{r=0}^k\frac{k!}{r!}\,x^{2r}\,,$$
which in turn implies that
$$\begin{align*}
2^{n+1}e^{x^2}\int_x^\infty e^{-t^2}t^{n+1}P_n(x/t)\,dt=&\,2^{n+1}e^{x^2}\int_x^\infty e^{-t^2}\frac1{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}t^{2k+1}\,dt\\[2mm]
=&\,2^{n+1}e^{x^2}\frac1{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}\underbrace{\int_x^\infty e^{-t^2}t^{2k+1}\,dt}_{=A_k}\\[2mm]
=&\,2e^{x^2}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}\,\frac{e^{-x^2}}2\sum_{r=0}^k\frac{k!}{r!}\,x^{2r}\\[2mm]
=&\,\sum_{k=0}^{\lfloor n/2\rfloor}\sum_{r=0}^k\frac{(-1)^k(2n-2k)!}{k!(n-k)!(n-2k)!}\,x^{n-2k}\,\frac{k!}{r!}\,x^{2r}\\[2mm]
=&\,\sum_{k=0}^{\lfloor n/2\rfloor}\sum_{r=0}^k\frac{(-1)^k(2n-2k)!}{r!(n-k)!(n-2k)!}\,x^{n-2(k-r)}\,.
\end{align*}$$
Our objective is to show that
$$\sum_{m=0}^{\lfloor n/2\rfloor}\frac{(-1)^mn!2^{n-2m}}{m!(n-2m)!}\,x^{n-2m}=\sum_{k=0}^{\lfloor n/2\rfloor}\sum_{r=0}^k\frac{(-1)^k(2n-2k)!}{r!(n-k)!(n-2k)!}\,x^{n-2(k-r)}\,.$$
Note that when $k$ and $r$ ranges as in the right-hand side double sum above, then the quantity $k-r$ varies precisely on the set $\bigl\{0,1,\dots,\lfloor n/2\rfloor\bigr\}$. Given $m$ in this set, then the possible pairs $(k,r)$ with $k-r=m$ are precisely those satisfying $m\leq k\leq\lfloor n/2\rfloor$ and $r=k-m$, so the right-hand side sum above can be rewritten as
$$\begin{align*}
&\,\sum_{m=0}^{\lfloor n/2\rfloor}\,\Biggl[\,\sum_{k=m}^{\lfloor n/2\rfloor}\frac{(-1)^k(2n-2k)!}{(k-m)!(n-k)!(n-2k)!}\Biggr]\,x^{n-2m}\\[2mm]
=&\,\sum_{m=0}^{\lfloor n/2\rfloor}\,\Biggl[\,\sum_{k=m}^{\lfloor n/2\rfloor}(-1)^k\,\binom{2n-2k}n\,\binom nk\,\binom km\,m!\,\Biggr]\,x^{n-2m}\,,
\end{align*}$$
so it remains to prove that for any $n,m$ with $0\leq m\leq\lfloor n/2\rfloor$ the following equality holds:
$$\frac{(-1)^mn!2^{n-2m}}{m!(n-2m)!}=\sum_{k=m}^{\lfloor n/2\rfloor}(-1)^k\,\binom{2n-2k}n\,\binom nk\,\binom km\,m!\,,$$
or, equivalently,
$$(-1)^m2^{n-2m}\,\binom n{2m}\,\binom{2m}m=\sum_{k=0}^n(-1)^k\,\binom{2n-2k}n\,\binom nk\,\binom km\tag{$\boldsymbol{\ast}$}$$
(the summands with $k<m$ or $\lfloor n/2\rfloor<k\leq n$ are equal to $0$).
I already have a proof of this fact, but it is too long to write down, so I will include later. This proves the result.
ADDENDUM
Let $n$ and $m$ be as before, fixed. Let $f(z)=\sum_{j=0}^\infty(-1)^j\,\binom nj\,\binom jm\,z^j$ and $g(z)=\sum_{j=0}^\infty\binom{2j}n\,z^j$. Then the right-hand side of $\boldsymbol{(\ast)}$ is precisely the coefficient of $z^n$ in the power series expansion of $f(z)g(z)$. I use Mathematica to obtain the formulas for $f$ and $g$, and afterwards I $\ $ construct $\ \ $ try to construct a direct proof of the equalities (thanks Mathematica!). Now
$$\begin{align*}
f(z)=&\,\sum_{j=m}^n(-1)^j\,\frac{n!j!}{j!(n-j)!m!(j-m)!}\,\frac{(n-m)!}{(n-m)!}\,z^j\\[2mm]
=&\,\sum_{j=m}^n(-1)^j\,\binom{n-m}{j-m}\,\binom nm\,z^j\\[2mm]
=&\,\sum_{j=0}^{n-m}(-1)^{j+m}\,\binom{n-m}j\,\binom nm\,z^{j+m}\\[2mm]
=&\,\binom nm\,(-z)^m(1-z)^{n-m}\,.
\end{align*}$$
On the other hand, denote by $\lceil x\rceil$ the least integer greater or equal than $x$. Then
$$g(z)=\sum_{\substack{j\geq0\\2j\geq n}}\binom{2j}n\,z^j=\sum_{j=\lceil n/2\rceil}^\infty\binom{2j}n\,z^j\,.$$
If $k=2j-n$ then $k\equiv n(\bmod\ 2)$, and $k\geq0$ iff$j\geq\lceil n/2\rceil$. Therefore the sum above can be rewritten as
$$\begin{align*}
g(z)=&\,\sum_{\substack{k\geq0\\k\equiv n(\bmod\ 2)}}\binom{k+n}n\,z^{\,(k+n)/2}\\
=&\,\frac{(-1)^n}2\sum_{k=0}^\infty\binom{k+n}n\,\bigl[(-1)^k-(-1)^{n+1}\bigr]\,z^{\,(k+n)/2}\\
=&\,\frac{(-1)^nz^{n/2}}2\sum_{k=0}^\infty\binom{k+n}n\,\bigl[(-1)^k-(-1)^{n+1}\bigr]\,z^{\,k/2}\\
=&\,\frac{(-1)^nz^{n/2}}2\sum_{k=0}^\infty\binom{k+n}n\,\bigl[\bigl(-\sqrt z\bigr)^k-(-1)^{n+1}\bigr(\sqrt z\bigr)^k\bigr]\,.\\
\end{align*}$$
Now we use the fact that $(1-\alpha)^{-(n+1)}=\sum_{k=0}^\infty\binom{k+n}k\,\alpha^k=\sum_{k=0}^\infty\binom{k+n}n\,\alpha^k$ (for $|\alpha|$ small), obtaining
$$\begin{align*}
g(z)=&\,\frac{(-1)^nz^{n/2}}2\Biggl[\frac1{\ \ \bigl[1-\bigl(-\sqrt z\bigr)\bigr]^{\,n+1}}-(-1)^{n+1}\frac1{\ \ \bigl[1-\sqrt z\,\bigr]^{\,n+1}}\Biggr]\\
=&\,\frac{(-1)^nz^{n/2}}2\,\frac{\bigl(\sqrt z-1\bigr)^{\,n+1}-\bigl(\sqrt z+1\bigr)^{\,n+1}}{(z-1)^{n+1}}\,.
\end{align*}$$
Therefore
$$\begin{align*}
f(z)g(z)=&\,\binom nm\,(-z)^m(1-z)^{n-m}\,\frac{(-1)^nz^{n/2}}2\,\frac{\bigl(\sqrt z-1\bigr)^{\,n+1}-\bigl(\sqrt z+1\bigr)^{\,n+1}}{(z-1)^{n+1}}\\
=&\,\binom nm\,z^m\,\frac{z^{n/2}}2\,\frac{\bigl(\sqrt z-1\bigr)^{\,n+1}-\bigl(\sqrt z+1\bigr)^{\,n+1}}{(z-1)^{m+1}}\\
=&\,\binom nm\,\frac{z^m}{2(z-1)^{m+1}}\,z^{n/2}\ \sum_{k=0}^{n+1}\binom{n+1}k\,\bigl(\sqrt z\bigr)^{\,k}\bigl[(-1)^{n+1-k}-1\bigr]\\
=&\,\binom nm\,\frac{z^m}{(z-1)^{m+1}}\sum_{\substack{0\leq k\leq n+1\\k\equiv n(\bmod\ 2)}}-\binom{n+1}k\,z^{\,(k+n)/2}\\
=&\,(-1)^m\binom nm\,z^m\Biggl[\sum_{\substack{0\leq k\leq n+1\\k\equiv n(\bmod\ 2)}}\binom{n+1}k\,z^{\,(k+n)/2}\Biggr]\sum_{r=0}^\infty\binom{r+m}m\,z^r\,.
\end{align*}$$
Taking $k=2j-n$, we see that $0\leq k\leq n+1$ iff $\frac n2\leq j\leq n+\frac12$, which implies
$$\begin{align*}
f(z)g(z)=&\,(-1)^m\binom nm\,z^m\Biggl[\sum_{j=\lceil n/2\rceil}^n\binom{n+1}{2j-n}\,z^j\,\Biggr]\sum_{r=0}^\infty\binom{r+m}m\,z^r\\
=&\,(-1)^m\binom nm\,z^m\Biggl[\sum_{j=0}^\infty\binom{n+1}{2j-n}\,z^j\,\Biggr]\sum_{r=0}^\infty\binom{r+m}m\,z^r\\
\end{align*}$$
It remains to show that the coefficient of $z^{n-m}$ in $\Bigl[\sum_{j=0}^\infty\binom{n+1}{2j-n}\,z^j\,\Bigr]\sum_{r=0}^\infty\binom{r+m}m\,z^r$ is equal to $2^{n-2m}\binom{n-m}m$. Unfortunately, I was unable to prove this, but Mathematica says that it is true: taking $n=2\ell$ the software confirms that
$$2^{2(\ell-m)}\,\binom{2\ell-m}m-\sum_{j=\ell}^{2\ell-m}\binom{2\ell+1}{2(j-\ell)}\,\binom{2\ell-j}m=0\,,$$
and similarly it is confirmed that for $n=2\ell+1$ the equality
$$2^{2(\ell-m)+1}\,\binom{2\ell+1-m}m-\sum_{j=\ell+1}^{2\ell+1-m}\binom{2(\ell+1)}{2(j-\ell)-1}\,\binom{2\ell+1-j}m=0$$
holds.
Best Answer
I might as well: there is an algorithm, due to Salzer, for converting to and from different orthogonal polynomial expansions. (I previously talked about this algorithm here.) If you intend to do this entire business of conversions, I believe an algorithm might be more useful to you than an integral expression.
What follows are a few relevant Mathematica routines. Even though I used Mathematica, I hope that the algorithm is still transparent and easily translatable (the indexing starts from 1; some adjustment is necessary if the indexing in your language starts at 0):
One could use Salzer again for converting from the Legendre to the monomial basis. However, it is a bit clearer (and gives rise to essentially the same method) to treat the problem as one of Taylor/Maclaurin expansion of the polynomial expressed in terms of Legendre polynomials. One can then use Clenshaw's algorithm for the purpose:
As an example, to demonstrate the identity
$$5P_0(x)-4P_1(x)-2P_2(x)+3P_3(x)+P_4(x)=\frac{35}{8}x^4+\frac{15}{2}x^3-\frac{27}{4}x^2-\frac{17}{2}x+\frac{51}{8}$$