# Is it possible to show a double integral from the product of two?

I have two identical integrals here. If I multiply them by each other is it possible that a double integral will result ?

1) Integrate[t^(x - 1)/E^t, {t, 0, Infinity}]*Integrate[s^(x - 1)/E^s, {s, 0, Infinity}]

ConditionalExpression[Gamma[x]^2, Re[x] > 0]

2)

Integrate[(t^(x - 1)*s^(x - 1))/E^(t + s), {s, 0, Infinity}, {t, 0, Infinity}]

ConditionalExpression[Gamma[x]^2, Re[x] > 0]

1. ..not derived with MMA

$$\int_0^{\infty } e^{-s} s^{x-1} \, ds \left(\int_0^{\infty } e^{-t} t^{x-1} \, dt\right)=\int _0^{\infty }\int _0^{\infty }e^{-(s+t)} s^{x-1} t^{x-1}dtds$$

Integrate[t^(x - 1)/E^t, {t, 0, Infinity}]*Integrate[s^(x - 1)/E^s, {s, 0, Infinity}] == Integrate[(t^(x - 1)*s^(x - 1))/E^(t + s), {s, 0, Infinity}, {t, 0, Infinity}]

ConditionalExpression[True, Re[x] > 0]

• As the integrand separates, the integrals can be evaluated separately and the individual results multiplied. Commented Apr 11, 2022 at 9:24
• @Daniel Huber, thanks The question is of MMA (can) show a double integral out of two separated integrals directly. Commented Apr 11, 2022 at 11:14
• That would not be very clever. A double integral is much more complicated than the product of 2 single integrals. Commented Apr 11, 2022 at 14:45
• I cannot figure out what exactly is the question. Are you asking whether a product of symbolic integrals will be transformed automatically to a multiple integral? If that is the question, the answer is no. Commented Apr 11, 2022 at 15:13
• @DanielLichtblau, thanks.That's too much asked then. I follow a prove and i know that the double integral equals the two single integrals So i can go further.. Commented Apr 11, 2022 at 15:35

If I understand correctly you just want to display it. In the following I used Inactive as in another recent answer to one of your questions. You might want to look up HoldForm and Defer for display purposes.

Relevant comment: though a nicer display might be aesthetically pleasing, I am not a big fan of it for practical purposes and long computations.

Define the following:

int[f_, g_, t1_, t2_, s1_, s2_] :=
Inactive[Integrate][f, {t, t1, t2}] Inactive[Integrate][
g, {s, s1, s2}] =
Inactive[Integrate][f g, {t, t1, t2}, {s, s1, s2}]


int[t^(x - 1)/E^t, s^(x - 1)/E^s, 0, Infinity, 0, Infinity]


If you want to calculate it now

Assuming[Re[x] > 0,
Activate@int[t^(x - 1)/E^t, s^(x - 1)/E^s, 0, Infinity, 0, Infinity]]


• bmf ,thanks. With this double integral i can try to follow the proof of Legendre's Duplicatie Formula ( known by MMA ?) Commented Apr 11, 2022 at 15:54
• @janhardo it's not clear to me what the additional question is to be honest. I just provided an answer for the question in the OP.
– bmf
Commented Apr 11, 2022 at 15:55
• bmf, the additional question has not been formulated yet, but will pop up as I follow the proof further regarding integration techniques. Commented Apr 11, 2022 at 15:59
• @janhardo perhaps you'd be interested in knowing that on SE in general each post should be self-contained as well as pose a single question. The reasoning is fairly simple in my eyes, if you keep adding stuff as questions and sub-questions a post will never be completed. You might want to keep that in mind
– bmf
Commented Apr 11, 2022 at 16:01
• bmf , Good that you pointed that out to me I then ask a new question about the topic I am working on then in a new post. Commented Apr 11, 2022 at 16:06

Double integral shows the same result as the product of singl integrals

{ Integrate[t^(x - 1)/E^t, {t, 0, Infinity}]*Integrate[s^(x - 1)/E^s,  {s, 0,Infinity}],
Integrate[t^(x - 1)/E^t s^(x - 1)/E^s, {s, 0, Infinity}, {t, 0, Infinity}]}

(*{ConditionalExpression[Gamma[x]^2, Re[x] > 0],
ConditionalExpression[Gamma[x]^2, Re[x] > 0]}*)

• thanks . It is clear to me that this is equal Can MMA symbolically make a double integral out of 2 integrals, that's really the question. Commented Apr 11, 2022 at 11:17