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Evaluate for $x\ge0$ $$F(x):=\int_0^1 \frac{t^x-1}{\log t}dt$$

After some testing I got that $F(x)$ should be $\log(x+1)$, but I can't prove it quite. My first try was to substitute $k:=\log t \Rightarrow t=e^k \Rightarrow dt=e^k dk$ and we get $F(x)=\int_{-\infty}^0 \frac{e^{kx}-1}{k}e^kdk=\int_0^{\infty}\frac{1-e^{-kx}}{k}e^{-k}dk$. I have had only functions where x is in the interval of integration. A hint on how to solve such problems would be nice.

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  • $\begingroup$ Differentiate with respect to $t$ and integrate with respect to $x$. $\endgroup$ – Jack Tiger Lam Aug 30 '16 at 21:07
  • $\begingroup$ @JackLam You mean I should differentiate $\frac{t^x-1}{\log t}$ with respect to t and then intigrate with respect to x? $\endgroup$ – HeatTheIce Aug 30 '16 at 21:21
  • $\begingroup$ ...Yes, my bad... $\endgroup$ – Jack Tiger Lam Aug 30 '16 at 21:22
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    $\begingroup$ You might like to read this fy.chalmers.se/~tfkhj/FeynmanIntegration.pdf $\endgroup$ – David Quinn Aug 30 '16 at 21:24
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$${ F }^{ \prime }\left( x \right) =\int _{ 0 }^{ 1 }{ \frac { \partial }{ \partial x } } \left( \frac { t^{ x }-1 }{ \log t } \right) dt=\int _{ 0 }^{ 1 }{ { t }^{ x }dx } =\frac { 1 }{ x+1 } $$ $$F\left( x \right) =\log { \left( x+1 \right) } +C$$ $$F\left( 0 \right) =\log { \left( 0+1 \right) } +C\\ C=0$$ so $$F(x)=\int _{ 0 }^{ 1 } \frac { t^{ x }-1 }{ \log t } dt=\color{red}{\log { \left( x+1 \right) } }$$

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Note that $\frac{t^x-1}{\log(t)}=\int_0^x t^{y}\,dy$. Therefore, we have

$$\begin{align} \int_0^1 \frac{t^x-1}{\log(t)}\,dt&=\int_0^1 \int_0^x t^{y}\,dy\,dt\\\\ &=\int_0^x \int_0^1 t^y\,dt\,dy\\\\ &=\int_0^x \frac{1}{y+1}\,dx\\\\ &=\log(x+1) \end{align}$$

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