Evaluate the definite integral $\int _0^1x^bb^x\,dx$.

Question

Since, the question asks to evaluate $\displaystyle \int \limits _0^1x^bb^x\,dx$. Sharing my thought shots on the same:

I multiplied and divided my integrand with $\log b^x$ and afterwards substituted $b^x=t$. So my Integrand reduced to $\displaystyle \int \limits _1^b\frac{(\log t)^b}{(\log b)^{b+1}}\,dt$. I can't see things getting better ahead. If someone can help!
Thanks in advance.
Please note that this is not the case with $b\to \infty$.

Does this help https://math.stackexchange.com/questions/4448712/finding-the-value-of-lim-a-to-infty-int-01-ax-xa-dx ? — Koro, Jun 02 '22 at 14:30
Do you know the technique/ integration and differentiation under the integral sign? — Shimura Variety, Jun 02 '22 at 14:34
@Koro, since the actual question that I had asked wasn't an improper integral but when I reduced it became one. Therefore, I can link both the questions for the time being. — Aastha Choudhary, Jun 02 '22 at 14:42
Please note that after substitution, the integral limits in OP should be from $1$ to $b$ and not from $0$ to $1$. — Koro, Jun 02 '22 at 14:47
@Koro Why was the question closed as the mentioned question only finds the limit as $x\rightarrow \infty$? Is it unsolvable in terms of elementary functions? — Shimura Variety, Jun 02 '22 at 14:55
@DevanshBhardwaj: It seems that the OP wanted to find the limit of the integral as $b\to \infty$ because they accepted in the comment that the linked post helped them. — Koro, Jun 02 '22 at 15:01
@Koro I am not sure as the user only says for the time being. Also, the user says is that it was not an improper integral, so please consider reopening the question. Maybe if we can answer in the general form, it can be helpful — Shimura Variety, Jun 02 '22 at 15:05
@AasthaChoudhary see that with the substitution $\log t=\xi$, the integrand becomes \begin{align}\frac{1}{(\log b)^{b+1}}\int_{0}^{\ln b}\xi^ne^{\xi}d\xi\end{align} This can be handled by a recursive formula — Shimura Variety, Jun 02 '22 at 15:15
@DevanshBhardwaj please mention that formula. The same recursion was happening when I was applying by-parts without substituting log t = A. — Aastha Choudhary, Jun 02 '22 at 15:29
Actually (i) The recursion with log t is different as upon differentiating it would lead to x coming into the denominator, which will complicate it. (ii) I am inviting you to chat as comments are not for extended discussion — Shimura Variety, Jun 02 '22 at 15:33
@AasthaChoudhary join this room :https://chat.stackexchange.com/rooms/136770/discussion-of-an-integral — Shimura Variety, Jun 02 '22 at 15:36
@DevanshBhardwaj I am not permitted for discussions since I don't have reputations more than 20. — Aastha Choudhary, Jun 02 '22 at 15:43

Тyma Gaidash · Answer 1 · 2022-06-03T16:56:56.853

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This integral looks like a gamma function:

$$\int_0^1 b^x x^b dx=\int_0^1 e^{\ln(b)x}x^b dx$$

Take $-\frac y{\ln(b)}=x$:

$$\int_0^{-\ln(b)} e^{-y} \left(-\frac y{\ln(b)}\right)^b\cdot -dy\frac1{\ln(b)}=(-\ln(b))^{-b-1}\int_0^{-\ln(b)} e^{-y} y^b dy$$

Remember the Incomplete Gamma function $\gamma(a,x)=\int_0^x k^{a-1} e^{-k}dk$:

$$(-\ln(b))^{-b-1}\int_0^{-\ln(b)} e^{-y} y^b dy =\frac{γ(b+1,-\ln(b))}{(-\ln(b))^{b+1}}$$

Here is a tester of the result which works for all $b>0$, but also works for some $b\in\Bbb C$

edited Jun 03 '22 at 16:56

answered Jun 03 '22 at 03:00

Тyma Gaidash

12,081

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Would it not be $(-1)^{b+1}$ as $dk=-dy$? – Shimura Variety Jun 03 '22 at 04:09
For $b>1$, would $(-1)^{b+1}$ not be imaginary for non integral b – Shimura Variety Jun 03 '22 at 04:19
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@DevanshBhardwaj Do you mean non integer $b$?. Also edited. – Тyma Gaidash Jun 03 '22 at 15:44
yeah greater than 1 – Shimura Variety Jun 03 '22 at 15:45
@DevanshBhardwaj The error was in $(-\ln(b))^b\ne (-1)^b \ln(b)^b$ for $b\in\Bbb R$, but the answer is now improved – Тyma Gaidash Jun 03 '22 at 16:46
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Well if b>1, then $-\ln b=\ln \frac{1}{b}$, but it is negative and so the term rised to the power b+1 is still negative. In particular,
$(-1)\ln b= (0-1)\ln b=0-\ln b =\ln (1)- \ln (b)=\ln (1/b)=-\ln b$, so there is really nothing wrong with stating it either way as they both are equivalent
– Shimura Variety Jun 03 '22 at 17:22
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Well actually I was missing the fact that $y^b$ would be imaginary for $b>1$ – Shimura Variety Jun 05 '22 at 10:41

Shimura Variety · Accepted Answer · 2022-06-05T10:42:51.703

After getting the integral as: \begin{align}\int_{1}^{b}\frac{(\ln t)^b}{(\ln b)^{b+1}}\,dx\end{align} It can be seen that upon using the substitution $\ln t = \xi$, the integral gets transformed into: \begin{align}\frac{1}{(\ln b)^{b+1}}\int_{0}^{\ln b}\xi^be^{\xi}\,d\xi\end{align} If b is a positive integer, then a recursive method is applicable. If b is a number which is not an integer, we transform the integral \begin{align}\frac{1}{(\ln b)^{b+1}}\int_{0}^{-\ln \frac{1}{b}}\xi^be^{\xi}\,d\xi=-\frac{(-1)^b}{(\ln b)^{b+1}}\int_{0}^{\ln \frac{1}{b}}\xi^be^{-\xi}\,d\xi\end{align} \begin{align}=\frac{(-1)^{b+1}}{(\ln b)^{b+1}}\int_{0}^{\ln \frac{1}{b}}\xi^be^{-\xi}\,d\xi=\frac{1}{(\ln \frac{1}{b})^{b+1}}\int_{0}^{\ln \frac{1}{b}}\xi^be^{-\xi}\,d\xi\end{align} Further, the Incomplete Gamma Function can be used to give $$ (\ln (\frac{1}{b}))^{-b-1}\gamma(b+1,\ln (\frac{1}{b})) $$ Coming to the recursion formula, suppose $f(x)=\xi^b$ and $g(x)=e^{\xi}$. Then we know that $g'(x)=g(x)$ This answer explains the recursion, so the solution is \begin{align} \frac{1}{(\ln b)^{b+1}}\bigg[\sum\limits_{k = 0}^b {( - 1)^{b - k} \frac{{b!}}{{k!}}\xi^k } \bigg]e^{\xi}\Biggr|_{0}^{\ln b} \end{align}

Evaluate the definite integral $\int _0^1x^bb^x\,dx$.

2 Answers2