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We have $f(m)>0.53899$ for $m$ sufficiently large. Under the Riemann hypothesis, we even have $f(m)>0.69314$ for $m$ sufficiently large, which would also imply that $f(m)$ attains a minimum.

If $m-1$ is a prime, then clearly $f(m)\geq 1$. Otherwise we have $$f(m)=\int_0^{m-1}\frac{d(\pi(x)-\pi(m))}{m-x}=\frac{\pi(m)}{m}+\int_0^{m-1}\frac{\pi(m)-\pi(x)}{(m-x)^2}\,dx.$$ Fix a constant $7/12<c<1$. By a result of Huxley (1972), we have $$\pi(m)-\pi(x)\geq(1+o(1))\frac{m-x}{\log(m-x)}\qquad\text{for}\qquad x<m-m^c.$$ Here, $o(1)$ is meant as $m$ tends to infinity. Therefore, $$f(m)\geq(1+o(1))\int_0^{m-m^c}\frac{1}{(m-x)\log(m-x)}\,dx=\log(1/c)+o(1).$$ As $\log(12/7)>0.53899$, my first claim is proved. Under the Riemann hypothesis, we can take any $1/2<c<1$, and then $\log 2>0.69314$ justifies my second claim.

Added. By a variant of the above argument, one can get an explicit lower bound for all $m$ (rather than for $m>m_0$) using an explicit prime number theorem in short intervals in place of Huxley's result. For example, one can use the work of Dudek (as explained in this earlier MO post of mine) to get $$ \sum_{x<p\leq x+3x^{2/3}}\log p>0.0006x^{2/3},\qquad x>\exp(8\times 10^{14}).$$

We have $f(m)>0.53899$ for $m$ sufficiently large. Under the Riemann hypothesis, we even have $f(m)>0.69314$ for $m$ sufficiently large, which would also imply that $f(m)$ attains a minimum.

If $m-1$ is a prime, then clearly $f(m)\geq 1$. Otherwise we have $$f(m)=\int_0^{m-1}\frac{d(\pi(x)-\pi(m))}{m-x}=\frac{\pi(m)}{m}+\int_0^{m-1}\frac{\pi(m)-\pi(x)}{(m-x)^2}\,dx.$$ Fix a constant $7/12<c<1$. By a result of Huxley (1972), we have $$\pi(m)-\pi(x)\geq(1+o(1))\frac{m-x}{\log(m-x)}\qquad\text{for}\qquad x<m-m^c.$$ Here, $o(1)$ is meant as $m$ tends to infinity. Therefore, $$f(m)\geq(1+o(1))\int_0^{m-m^c}\frac{1}{(m-x)\log(m-x)}\,dx=\log(1/c)+o(1).$$ As $\log(12/7)>0.53899$, my first claim is proved. Under the Riemann hypothesis, we can take any $1/2<c<1$, and then $\log 2>0.69314$ justifies my second claim.

Added. By a variant of the above argument, one can get an explicit lower bound for all $m$ (rather than for $m>m_0$) using an explicit prime number theorem in short intervals (in place of Huxley's result). For example, one can use the work of Dudek (as explained in this earlier MO post of mine) to get $$ \sum_{x<p\leq x+3x^{2/3}}\log p>0.0006x^{2/3},\qquad x>\exp(8\times 10^{14}).$$

We have $f(m)>0.53899$ for $m$ sufficiently large. Under the Riemann hypothesis, we even have $f(m)>0.69314$ for $m$ sufficiently large, which would also imply that $f(m)$ attains a minimum.

If $m-1$ is a prime, then clearly $f(m)\geq 1$. Otherwise we have $$f(m)=\int_0^{m-1}\frac{d(\pi(x)-\pi(m))}{m-x}=\frac{\pi(m)}{m}+\int_0^{m-1}\frac{\pi(m)-\pi(x)}{(m-x)^2}\,dx.$$ Fix a constant $7/12<c<1$. By a result of Huxley (1972), we have $$\pi(m)-\pi(x)\geq(1+o(1))\frac{m-x}{\log(m-x)}\qquad\text{for}\qquad x<m-m^c.$$ Here, $o(1)$ is meant as $m$ tends to infinity. Therefore, $$f(m)\geq(1+o(1))\int_0^{m-m^c}\frac{1}{(m-x)\log(m-x)}\,dx=\log(1/c)+o(1).$$ As $\log(12/7)>0.53899$, my first claim is proved. Under the Riemann hypothesis, we can take any $1/2<c<1$, and then $\log 2>0.69314$ justifies my second claim.

We have $f(m)>0.53899$ for $m$ sufficiently large. Under the Riemann hypothesis, we even have $f(m)>0.69314$ for $m$ sufficiently large, which would also imply that $f(m)$ attains a minimum.

If $m-1$ is a prime, then clearly $f(m)\geq 1$. Otherwise we have $$f(m)=\int_0^{m-1}\frac{d(\pi(x)-\pi(m))}{m-x}=\frac{\pi(m)}{m}+\int_0^{m-1}\frac{\pi(m)-\pi(x)}{(m-x)^2}\,dx.$$ Fix a constant $7/12<c<1$. By a result of Huxley (1972), we have $$\pi(m)-\pi(x)\geq(1+o(1))\frac{m-x}{\log(m-x)}\qquad\text{for}\qquad x<m-m^c.$$ Here, $o(1)$ is meant as $m$ tends to infinity. Therefore, $$f(m)\geq(1+o(1))\int_0^{m-m^c}\frac{1}{(m-x)\log(m-x)}\,dx=\log(1/c)+o(1).$$ As $\log(12/7)>0.53899$, my first claim is proved. Under the Riemann hypothesis, we can take any $1/2<c<1$, and then $\log 2>0.69314$ justifies my second claim.

Added. By a variant of the above argument, one can get an explicit lower bound for all $m$ (rather than for $m>m_0$) using an explicit prime number theorem in short intervals in place of Huxley's result. For example, one can use the work of Dudek (as explained in this earlier MO post of mine) to get $$ \sum_{x<p\leq x+3x^{2/3}}\log p>0.0006x^{2/3},\qquad x>\exp(8\times 10^{14}).$$

We have $f(m)>0.53899$ for $m$ sufficiently large. Under the Riemann hypothesis, we even have $f(m)>0.69314$ for $m$ sufficiently large, which would also imply that $f(m)$ attains a minimum.

If $m-1$ is a prime, then clearly $f(m)\geq 1$. Otherwise we have $$f(m)=\int_0^{m-1}\frac{d(\pi(x)-\pi(m))}{m-x}=\frac{\pi(m)}{m}+\int_0^{m-1}\frac{\pi(m)-\pi(x)}{(m-x)^2}\,dx.$$ Fix a constant $7/12<c<1$. By a result of Huxley (1972), we have $$\pi(m)-\pi(x)\geq(1+o(1))\frac{m-x}{\log(m-x)}\qquad\text{for}\qquad x<m-m^c.$$ Therefore, $$f(m)\geq(1+o(1))\int_0^{m-m^c}\frac{1}{(m-x)\log(m-x)}\,dx=\log(1/c)+o(1).$$ As $\log(12/7)>0.53899$, my first claim is proved. Under the Riemann hypothesis, we can take any $1/2<c<1$, and then $\log 2>0.69314$ justifies my second claim.

We have $f(m)>0.53899$ for $m$ sufficiently large. Under the Riemann hypothesis, we even have $f(m)>0.69314$ for $m$ sufficiently large, which would also imply that $f(m)$ attains a minimum.

If $m-1$ is a prime, then clearly $f(m)\geq 1$. Otherwise we have $$f(m)=\int_0^{m-1}\frac{d(\pi(x)-\pi(m))}{m-x}=\frac{\pi(m)}{m}+\int_0^{m-1}\frac{\pi(m)-\pi(x)}{(m-x)^2}\,dx.$$ Fix a constant $7/12<c<1$. By a result of Huxley (1972), we have $$\pi(m)-\pi(x)\geq(1+o(1))\frac{m-x}{\log(m-x)}\qquad\text{for}\qquad x<m-m^c.$$ Here, $o(1)$ is meant as $m$ tends to infinity. Therefore, $$f(m)\geq(1+o(1))\int_0^{m-m^c}\frac{1}{(m-x)\log(m-x)}\,dx=\log(1/c)+o(1).$$ As $\log(12/7)>0.53899$, my first claim is proved. Under the Riemann hypothesis, we can take any $1/2<c<1$, and then $\log 2>0.69314$ justifies my second claim.