probability that chi-squared random variable is less than other chi-squared random variable

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I have random variables $X_i = \mathcal N(0,\sigma_1^2)$ and $Y_i = \mathcal N(0,\sigma_2^2) $ where i = 0, 1 ,..., n. all of them are independent to one another

I obtained random variable $X = X_1^2 + X_2^2 + ... X_n^2$ and $Y=Y_1^2+Y_2^2...Y_n^2$ are chi-squared random variables with n degrees of freedom

I want to find the probability $P(X<Y)$, I can find the $f_X(x)$ and $f_Y(y)$

but i can not calculate $P(X<Y) = \int_{0}^{\infty}P(X<y)f_Y(y)dy$ because $P(X<y)$ and $f_Y(y)$ are complicated

Is there other method to find $P(X<Y)$ without above integral method? I really want to calculated the probability $P(X<Y)$