By the Hardy-Hilbert's integral inequality, the author refers to the inequality \[ \int^\infty_0\int^\infty_0 \frac{f(x)g(y)}{x+y} dx dy 1\), \(\frac 1{p} + \frac 1{q} = 1\), \(\lambda > 2-\min\{p,q\}\), \(\alpha < T \leq \infty\), \(f(t) \geq 0\), \(g(t) \geq 0\), \(0 < \int_\alpha ^T (t-\alpha)^{1-\lambda} f^{p}(t) dt < \infty\) and \(0 < \int_\alpha ^T (t-\alpha)^{1-\lambda} g^{q}(t) dt < \infty\). Then (i) for \(T < \infty,\) we have \[ \begin{aligned} \int^T _\alpha \int^T_\alpha& \frac{f(x)g(y)}{(x+y-2\alpha)^{\lambda}} dx dy\\ &< \Big\{\int^T_{\alpha}\Big[k_{\lambda}(p)-\theta_{\lambda}(p)\Big(\frac{t-\alpha} {T-\alpha}\Big)^{(p+\lambda-2)/p}\Big](t-\alpha)^{1-\lambda}f^{p}(t)dt\Big\}^{1/p}\\ &\times\Big\{\int^T_{\alpha}\Big[k_{\lambda}(p)-\theta_{\lambda}(q)\Big(\frac{t-\alpha} {T-\alpha}\Big)^{(q+\lambda -2)/q}\Big](t-\alpha)^{1-\lambda}g^{q}(t)dt\Big\}^{1/q},\end{aligned} \] \[ k_{\lambda}(p) = B\Big(\frac{p+\lambda -2}{p},\frac{q+\lambda -2}{q}\Big) \] and \[ \theta_{\lambda}(r) = \int^1_0 \frac 1{(1+u)^{\lambda}}\Big(\frac 1{u}\Big)^ {(2-\lambda)/r} du, \quad r = p,q; \] (ii) for \(T = \infty,\) we have \[ \int^\infty_0\int^\infty_0 \frac{f(x)g(y)}{(x+y-2\alpha)^{\lambda}} dx dy \] \[ < k_{\lambda}(p)\Big[\int^\infty_\alpha (t-\alpha)^{1-\lambda}f^{p}(t) dt\Big]^{1/p}\Big[\int^\infty_\alpha (t-\alpha)^{1-\lambda}g^{q}(t) dt\Big]^{1/q}, \] where the constants \(k_{\lambda}(p)\) is the best possible. The above inequalities are new improvements of the results of \textit{B. Yang} [J. Math. Anal. Appl. 220, No. 2, 778-785 (1998; Zbl 0911.26011)] and \textit{J. Kuang} [J. Math. Anal. Appl. 235, No. 2, 608-614 (1999; Zbl 0951.26013)]. Furthermore, the author points out that the constant \(k_{\lambda}(p)\) is the best possible and gives a more accurate estimate than his previous result [Acta Math. Sin. 41, No. 4, 839-844 (1998)].