Proposition (the Chebychev ψ function may be written as the sum of Chebyshev ϑ functions):
We have the identity
.
Proposition (estimate of the distance between the Chebychev ψ and ϑ functions):
Whenever
, we have
.
Note: The current proof gives an inferior error term. A subsequent version will redeem this issue. (Given the Riemann hypothesis, the error term can be made even smaller.)
Proof: We know that the formula

holds. Hence,
.
By a result obtained by Pierre Dusart (based upon the computational verification of the Riemann hypothesis for small moduli), we have

whenever
. If
is in that range, we hence conclude
.
By Euler's summation formula, we have
.
Certainly
and
. Moreover,
. Now derivation shows that

is an anti-derivative of the function

of
. By the fundamental theorem of calculus, it follows that
![{\displaystyle \int _{a}^{b}\left(1-{\frac {2t}{\ln(x)}}\right)x^{1/t}dt=\left[-\exp \left({\frac {\ln(x)}{t}}\right){\frac {t^{2}}{\ln(x)}}\right]_{t=a}^{t=b}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/807c2f3f7382b2ba026390291998197834e9c425)
for real numbers
such that
. This integral is not precisely the one we want to estimate. Hence, some analytical trickery will be necessary in order to obtain the estimate we want.
We start by noting that if only the bracketed term in the integral were absent, we would have the estimate we desire. In order to proceed, we replace
by the more general expression
(where
), and obtain
.
The integrand is non-negative so long as
.
Moreover, if
is strictly within that range, we obtain
.
We now introduce a constant
and obtain the integrals
and
.
The first integral majorises the integral
,
whereas the second integral majorises the integral
.
We obtain that
.
Now we would like to set
. To do so, we must ensure that
is sufficiently large so that
resp.
is strictly within the admissible interval.
The two summands on the left are now estimated using our computation above, where
is replaced by
for the first computation: Indeed,
![{\displaystyle \int _{2}^{K}x^{1/t}y_{1}^{1/t}dt\leq \left(1-{\frac {2K}{\ln(xy_{1})}}\right)^{-1}\left[-\exp \left({\frac {\ln(xy_{1})}{t}}\right){\frac {t^{2}}{\ln(xy_{1})}}\right]_{t=2}^{t=K}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62bbfb7e739b91e5719354a9105e544fa1af441f)
and
.
Putting the estimates together and setting
, we obtain
![{\displaystyle \int _{2}^{\log _{2}(x)}x^{1/t}dt\leq {\frac {1}{y_{1}^{1/K}}}\left(1-{\frac {2K}{\ln(xy_{1})}}\right)^{-1}\left[-\exp \left({\frac {\ln(xy_{1})}{t}}\right){\frac {t^{2}}{\ln(xy_{1})}}\right]_{t=2}^{t=K}+\left(1-{\frac {2t_{0}}{\ln(xy_{2})}}\right)^{-1}\left[-\exp \left({\frac {\ln(xy_{2})}{t}}\right){\frac {t^{2}}{\ln(xy_{2})}}\right]_{t=K}^{t=\log _{2}(x)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ee969c4ab5687cc5b7360da8742de22ba2ff0ea)
whenever
and
.
We now choose the ansatz
and 
for constants
and
. These equations are readily seen to imply
and
.
Note though that
and
is needed. The first condition yields
.
The equations for
and
may be inserted into the above constraints on
and
; this yields
and
, that is,
and
.
If all these conditions are true, the ansatz immediately yields
.
We now amend our ansatz by further postulating
.
This yields

and
.
From this we deduce that in order to obtain an asymptotically sharp error term, we need to set
. But doing so yields the desired result.